{"id":285,"date":"2014-12-11T02:30:08","date_gmt":"2014-12-11T02:30:08","guid":{"rendered":"https:\/\/courses.candelalearning.com\/colphysics\/?post_type=chapter&#038;p=285"},"modified":"2016-11-07T19:16:09","modified_gmt":"2016-11-07T19:16:09","slug":"2-5-motion-equations-for-constant-acceleration-in-one-dimension","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-physics\/chapter\/2-5-motion-equations-for-constant-acceleration-in-one-dimension\/","title":{"raw":"Motion Equations for Constant Acceleration in One Dimension","rendered":"Motion Equations for Constant Acceleration in One Dimension"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<div class=\"itemizedlist\">\r\n<ul class=\"itemizedlist\">\r\n \t<li class=\"listitem\">Calculate displacement of an object that is not accelerating, given initial position and velocity.<\/li>\r\n \t<li class=\"listitem\">Calculate final velocity of an accelerating object, given initial velocity, acceleration, and time.<\/li>\r\n \t<li class=\"listitem\">Calculate displacement and final position of an accelerating object, given initial position, initial velocity, time, and acceleration.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div id=\"m42099-import-auto-id1489960\" class=\"figure\" title=\"Figure 2.25.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"300\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101608\/Figure_02_05_00.jpg\" alt=\"Four men racing up a river in their kayaks.\" width=\"300\" height=\"404\" \/> Figure 1. Kinematic equations can help us describe and predict the motion of moving objects such as these kayaks racing in Newbury, England. (credit: Barry Skeates, Flickr)[\/caption]\r\n\r\nWe might know that the greater the acceleration of, say, a car moving away from a stop sign, the greater the displacement in a given time. But we have not developed a specific equation that relates acceleration and displacement. In this section, we develop some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration already covered.\r\n\r\n<\/div>\r\n<\/div>\r\n<h2><span class=\"cnx-gentext-section cnx-gentext-t\">Notation: <span class=\"emphasis\"><em>t<\/em><\/span>, <span class=\"emphasis\"><em>x<\/em><\/span>, <span class=\"emphasis\"><em>v<\/em><\/span>, <span class=\"emphasis\"><em>a<\/em><\/span><\/span><\/h2>\r\nFirst, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. Since elapsed time is <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>=\u00a0<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>f<\/sub>\u2212<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>0<\/sub><\/span>, taking <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>0\u00a0<\/sub>= 0<\/span> means that <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>=\u00a0<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>f<\/sub><\/span>, the final time on the stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity. That is, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>0<\/sub><\/span> <span class=\"emphasis\"><em>is the initial position<\/em><\/span> and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0<\/sub><\/span> <span class=\"emphasis\"><em>is the initial velocity<\/em><\/span>. We put no subscripts on the final values. That is, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> <span class=\"emphasis\"><em> is the final time<\/em><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> <span class=\"emphasis\"><em> is the final position<\/em><\/span>, and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span> <span class=\"emphasis\"><em>is the final velocity<\/em><\/span>. This gives a simpler expression for elapsed time\u2014now, <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span>=<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span>. It also simplifies the expression for displacement, which is now <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>x\u00a0<\/em><\/span>=\u00a0<span class=\"emphasis mathml-mi\"><em>x<\/em><\/span>\u2212<span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>0<\/sub><\/span>. Also, it simplifies the expression for change in velocity, which is now <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>v\u00a0<\/em><\/span>=\u00a0<span class=\"emphasis mathml-mi\"><em>v<\/em><\/span>\u2212<span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0<\/sub><\/span>. <span style=\"color: #000000;\">To summarize, using the simplified notation, with the initial time taken to be zero,<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{cases}{\\Delta}{t} &amp;=&amp; t \\\\{\\Delta}{x} &amp;=&amp; x-{{x}_{0}}\\\\{\\Delta}{v} &amp;=&amp; v-{{v}_{0}}\\end{cases}[\/latex]<\/p>\r\nwhere <span class=\"emphasis\"><em>the subscript 0 denotes an initial value and the absence of a subscript denotes a final value<\/em><\/span> in whatever motion is under consideration.\r\n\r\nWe now make the important assumption that <span class=\"emphasis\"><em>acceleration is constant<\/em><\/span>. This assumption allows us to avoid using calculus to find instantaneous acceleration. Since acceleration is constant, the average and instantaneous accelerations are equal. That is,\r\n<div id=\"m42099-import-auto-id1819116\" class=\"equation\" title=\"Equation 2.29.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]\\bar{a }=a=\\text{ constant}[\/latex],<\/div>\r\n<\/div>\r\nso we use the symbol <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span> for acceleration at all times. Assuming acceleration to be constant does not seriously limit the situations we can study nor degrade the accuracy of our treatment. For one thing, acceleration <span class=\"emphasis\"><em>is<\/em><\/span> constant in a great number of situations. Furthermore, in many other situations we can accurately describe motion by assuming a constant acceleration equal to the average acceleration for that motion. Finally, in motions where acceleration changes drastically, such as a car accelerating to top speed and then braking to a stop, the motion can be considered in separate parts, each of which has its own constant acceleration.\r\n<div class=\"section textbox shaded\" title=\"Notation: t, x, v, a\">\r\n<h3><strong><span class=\"cnx-gentext-tip-t\">Solving for Displacement (<span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>) and Final Position (<span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>) from Average Velocity when Acceleration (<span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span>) is Constant<\/span><\/strong><\/h3>\r\n<div id=\"m42099-fs-id1164906424651\" class=\"note\">\r\n<div class=\"body\">\r\n\r\nTo get our first two new equations, we start with the definition of average velocity:\r\n<div id=\"m42099-import-auto-id1782525\" class=\"equation\" title=\"Equation 2.30.\">\r\n<div class=\"mediaobject\">\r\n<div id=\"import-auto-id1782525\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\bar{v}=\\frac{\\Delta x}{\\Delta t}[\/latex].<\/div>\r\n<div class=\"equation\" style=\"text-align: center;\" data-type=\"equation\"><\/div>\r\n<\/div>\r\n<\/div>\r\nSubstituting the simplified notation for <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> and <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> yields\r\n<div id=\"m42099-import-auto-id4145482\" class=\"equation\" title=\"Equation 2.31.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]\\bar{v}=\\frac{x-{x}_{0}}{t}[\/latex]<\/div>\r\n<div class=\"mediaobject\" style=\"text-align: center;\"><\/div>\r\n<\/div>\r\nSolving for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> yields\r\n<div id=\"m42099-import-auto-id2297251\" class=\"equation\" title=\"Equation 2.32.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]x={x}_{0}+\\bar{v}t[\/latex],<\/div>\r\n<div class=\"mediaobject\" style=\"text-align: center;\"><\/div>\r\n<\/div>\r\nwhere the average velocity is\r\n<div class=\"MathJax_Display\" style=\"text-align: center;\">[latex]\\bar{v}=\\frac{{v}_{0}+v}{2}\\left(\\text{ constant }a\\right)[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"section\" title=\"Putting Equations Together\">\r\n<div class=\"titlepage\">\r\n\r\nThe equation [latex]\\bar{v}=\\frac{{v}_{0}+v}{2}[\/latex] reflects the fact that, when acceleration is constant, <em>v<\/em> is just the simple average of the initial and final velocities. For example, if you steadily increase your velocity (that is, with constant acceleration) from 30 to 60 km\/h, then your average velocity during this steady increase is 45 km\/h. Using the equation [latex]\\bar{v}=\\frac{{v}_{0}+v}{2}[\/latex]\u00a0to check this, we see that\r\n<div id=\"import-auto-id4152098\" data-type=\"equation\">\r\n<div class=\"MathJax_Display\" style=\"text-align: center;\">[latex]\\bar{v}=\\frac{{v}_{0}+v}{2}=\\frac{\\text{30 km\/h}+\\text{60 km\/h}}{2}=\\text{45 km\/h}[\/latex],<\/div>\r\n<\/div>\r\n<p id=\"import-auto-id2297106\">which seems logical.<\/p>\r\n\r\n<div id=\"m42099-fs-id1164906442368\" class=\"example\" title=\"Example 2.8. Calculating Displacement: How Far does the Jogger Run?\">\r\n<div class=\"textbox examples\">\r\n<h3>Example 1. Calculating Displacement: How Far does the Jogger Run?<\/h3>\r\nA jogger runs down a straight stretch of road with an average velocity of 4.00 m\/s for 2.00 min. What is his final position, taking his initial position to be zero?\r\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\r\nDraw a sketch.\r\n<div id=\"m42099-import-auto-id2297327\" class=\"figure\" title=\"Figure 2.26.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"400\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101613\/Figure_02_05_00a.jpg\" alt=\"Velocity vector arrow labeled v equals 4 point zero zero meters per second over an x axis displaying initial and final positions. Final position is labeled x equals question mark.\" width=\"400\" height=\"175\" \/> Figure 2.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\"><\/div>\r\n<\/div>\r\nThe final position <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> is given by the equation\r\n<div id=\"m42099-import-auto-id2168543\" class=\"equation\" title=\"Equation 2.35.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]x={x}_{0}+\\bar{v}t[\/latex].<\/div>\r\n<div class=\"mediaobject\" style=\"text-align: center;\"><\/div>\r\n<\/div>\r\nTo find <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>, we identify the values of <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>0<\/sub><\/span>,\u00a0[latex]\\bar{v}[\/latex], and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> from the statement of the problem and substitute them into the equation.\r\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\r\n<div id=\"m42099-import-auto-id2300752\" class=\"equation\" title=\"Equation 2.36.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">\r\n<p id=\"import-auto-id2169540\" style=\"text-align: left;\">1. Identify the knowns. [latex]\\bar{v}=4.00\\text{ m\/s}[\/latex], [latex]\\Delta t=2.00\\text{ min}[\/latex], and [latex]{x}_{0}=0\\text{ m}[\/latex].<\/p>\r\n<p id=\"import-auto-id2300750\" style=\"text-align: left;\">2. Enter the known values into the equation.<\/p>\r\n\r\n<div id=\"import-auto-id2300752\" class=\"equation\" data-type=\"equation\">[latex]x={x}_{0}+\\bar{v}t=0+\\left(4\\text{.}\\text{00 m\/s}\\right)\\left(\\text{120 s}\\right)=\\text{480 m}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\r\nVelocity and final displacement are both positive, which means they are in the same direction.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nThe equation [latex]x={x}_{0}+\\bar{v}t[\/latex]\u00a0gives insight into the relationship between displacement, average velocity, and time. It shows, for example, that displacement is a linear function of average velocity. (By linear function, we mean that displacement depends on\u00a0[latex]\\bar{v}[\/latex]<em data-effect=\"italics\">,<\/em>\u00a0rather than on\u00a0[latex]\\bar{v}[\/latex]\u00a0raised to some other power, such as\u00a0[latex]\\bar{v}^{2}[\/latex]. When graphed, linear functions look like straight lines with a constant slope.) On a car trip, for example, we will get twice as far in a given time if we average 90 km\/h than if we average 45 km\/h.\r\n<div id=\"m42099-import-auto-id1962019\" class=\"figure\" title=\"Figure 2.27.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"300\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101620\/Figure_02_05_00b.jpg\" alt=\"Line graph showing displacement in meters versus average velocity in meters per second. The line is straight with a positive slope. Displacement x increases linearly with increase in average velocity v.\" width=\"300\" height=\"446\" \/> Figure 3. There is a linear relationship between displacement and average velocity. For a given time t, an object moving twice as fast as another object will move twice as far as the other object.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\"><\/div>\r\n<\/div>\r\n<div id=\"m42099-fs-id1164906431223\" class=\"note\">\r\n<div class=\"title textbox shaded\">\r\n<h3 class=\"title\"><strong><span class=\"cnx-gentext-tip-t\">Solving for Final Velocity<\/span><\/strong><\/h3>\r\n<div class=\"body\">\r\n\r\nWe can derive another useful equation by manipulating the definition of acceleration.\r\n<div id=\"m42099-import-auto-id2366011\" class=\"equation\" title=\"Equation 2.37.\">\r\n<div class=\"mediaobject\">\r\n<div id=\"import-auto-id2366011\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]a=\\frac{\\Delta v}{\\Delta t}[\/latex]<\/div>\r\n<div class=\"equation\" style=\"text-align: center;\" data-type=\"equation\"><\/div>\r\n<\/div>\r\n<\/div>\r\nSubstituting the simplified notation for <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span> and <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> gives us\r\n<div id=\"m42099-import-auto-id1962571\" class=\"equation\" title=\"Equation 2.38.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]a=\\frac{v-{v}_{0}}{t}\\text{}\\left(\\text{constant}a\\right)[\/latex]<\/div>\r\n<div class=\"mediaobject\"><\/div>\r\n<\/div>\r\nSolving for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span> yields\r\n<div id=\"m42099-import-auto-id2168495\" class=\"equation\" title=\"Equation 2.39.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]v={v}_{0}+\\text{at}\\text{}\\left(\\text{constant}a\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"m42099-fs-id1164906431414\" class=\"example\" title=\"Example 2.9. Calculating Final Velocity: An Airplane Slowing Down after Landing\">\r\n<div class=\"textbox examples\">\r\n<h3><span class=\"cnx-gentext-example cnx-gentext-autogenerated\">Example <\/span><span class=\"cnx-gentext-example cnx-gentext-n\">2<\/span><span class=\"cnx-gentext-example cnx-gentext-autogenerated\">. <\/span><span class=\"cnx-gentext-example cnx-gentext-t\">Calculating Final Velocity: An Airplane Slowing Down after Landing<\/span><\/h3>\r\nAn airplane lands with an initial velocity of 70.0 m\/s and then decelerates at <span class=\"token\">1.50 m\/s<sup>2<\/sup><\/span> for 40.0 s. What is its final velocity?\r\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\r\nDraw a sketch. We draw the acceleration vector in the direction opposite the velocity vector because the plane is decelerating.\r\n<div id=\"m42099-import-auto-id2300963\" class=\"figure\" title=\"Figure 2.28.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"350\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101623\/Figure_02_05_00c.jpg\" alt=\"Velocity vector arrow pointing toward the right in the positive x direction. Initial velocity equals seventy meters per second. Final velocity equals question mark. An acceleration vector arrow pointing toward the left labeled a equals negative 1 point 50 meters per second squared.\" width=\"350\" height=\"209\" \/> Figure 4.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\r\n1. Identify the knowns. v<sub>0\u00a0<\/sub><span class=\"token\">= 70.0 m\/s<\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a\u00a0<\/em><\/span>= \u22121.50 m\/s<sup>2<\/sup><\/span>, <em>t<\/em> = 40.0 s.\r\n\r\n2. Identify the unknown. In this case, it is final velocity, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>f<\/sub><\/span><sub>.<\/sub>\r\n\r\n3. Determine which equation to use. We can calculate the final velocity using the equation [latex]v={v}_{0}+{at}[\/latex].\r\n\r\n4. Plug in the known values and solve.\r\n<div id=\"m42099-import-auto-id2177855\" class=\"equation\" title=\"Equation 2.40.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]v={v}_{0}+\\text{at}=\\text{70}\\text{.}\\text{0 m\/s}+\\left(-1\\text{.}{\\text{50 m\/s}}^{2}\\right)\\left(\\text{40}\\text{.}\\text{0 s}\\right)=\\text{10}\\text{.}\\text{0 m\/s}[\/latex]<\/div>\r\n<\/div>\r\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\r\nThe final velocity is much less than the initial velocity, as desired when slowing down, but still positive. With jet engines, reverse thrust could be maintained long enough to stop the plane and start moving it backward. That would be indicated by a negative final velocity, which is not the case here.\r\n<div id=\"m42099-import-auto-id2173965\" class=\"figure\" title=\"Figure 2.29.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"600\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101626\/Figure_02_04_01.jpg\" alt=\"An airplane moving toward the right at two points in time. At time equals 0 the velocity vector arrow points toward the right and is labeled seventy meters per second. The acceleration vector arrow points toward the left and is labeled negative 1 point 5 meters per second squared. At time equals forty seconds, the velocity arrow is shorter, points toward the right, and is labeled ten meters per second. The acceleration vector arrow is still pointing toward the left and is labeled a equals negative 1 point 5 meters per second squared.\" width=\"600\" height=\"202\" \/> Figure 5. The airplane lands with an initial velocity of 70.0 m\/s and slows to a final velocity of 10.0 m\/s before heading for the terminal. Note that the acceleration is negative because its direction is opposite to its velocity, which is positive.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nIn addition to being useful in problem solving, the equation [latex]v={v}_{0}+\\text{at}[\/latex]\u00a0gives us insight into the relationships among velocity, acceleration, and time. From it we can see, for example, that\r\n<div class=\"itemizedlist\">\r\n<ul class=\"itemizedlist\">\r\n \t<li class=\"listitem\">final velocity depends on how large the acceleration is and how long it lasts<\/li>\r\n \t<li class=\"listitem\">if the acceleration is zero, then the final velocity equals the initial velocity <span class=\"token\">(<span class=\"emphasis mathml-mi\"><em>v\u00a0<\/em><\/span>=\u00a0<span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0<\/sub>)<\/span>, as expected (i.e., velocity is constant)<\/li>\r\n \t<li class=\"listitem\">if <span class=\"emphasis\"><em><span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span><\/em><\/span> is negative, then the final velocity is less than the initial velocity<\/li>\r\n<\/ul>\r\n<\/div>\r\n(All of these observations fit our intuition, and it is always useful to examine basic equations in light of our intuition and experiences to check that they do indeed describe nature accurately.)\r\n<div id=\"m42099-fs-id1164906437247\" class=\"note\">\r\n<div class=\"textbox learning-objectives\">\r\n<h3><span class=\"cnx-gentext-tip-t\">Making Connections: Real-World Connection<\/span><\/h3>\r\n<section>\r\n<div data-type=\"note\">\r\n<div class=\"body\">\r\n\r\nAn intercontinental ballistic missile (ICBM) has a larger average acceleration than the Space Shuttle and achieves a greater velocity in the first minute or two of flight (actual ICBM burn times are classified\u2014short-burn-time missiles are more difficult for an enemy to destroy). But the Space Shuttle obtains a greater final velocity, so that it can orbit the earth rather than come directly back down as an ICBM does. The Space Shuttle does this by accelerating for a longer time.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<div class=\"title textbox shaded\">\r\n<div id=\"m42099-fs-id1164906444525\" class=\"note\">\r\n<h3 class=\"title\"><strong><span class=\"cnx-gentext-tip-t\">Solving for Final Position When Velocity is Not Constant (<span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a\u00a0<\/em><\/span>\u2260 0<\/span>)<\/span><\/strong><\/h3>\r\n<div class=\"body\">\r\n\r\nWe can combine the equations above to find a third equation that allows us to calculate the final position of an object experiencing constant acceleration. We start with\r\n<div id=\"m42099-import-auto-id2166975\" class=\"equation\" title=\"Equation 2.41.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]v={v}_{0}+{at}[\/latex]<\/div>\r\n<div class=\"mediaobject\" style=\"text-align: center;\"><\/div>\r\n<\/div>\r\nAdding <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0<\/sub><\/span> to each side of this equation and dividing by 2 gives\r\n<div id=\"m42099-import-auto-id1689620\" class=\"equation\" title=\"Equation 2.42.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]\\frac{{v}_{0}+v}{2}={v}_{0}+\\frac{1}{2}{at}[\/latex]<\/div>\r\n<div class=\"mediaobject\" style=\"text-align: center;\"><\/div>\r\n<\/div>\r\nSince [latex]\\frac{{v}_{0}+v}{2}=\\bar{v}[\/latex]\u00a0for constant acceleration, then\r\n<div id=\"m42099-import-auto-id2301379\" class=\"equation\" title=\"Equation 2.43.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]\\bar{v}={v}_{0}+\\frac{1}{2}{at}[\/latex]<\/div>\r\n<div class=\"mediaobject\"><\/div>\r\n<\/div>\r\nNow we substitute this expression for [latex]\\bar{v}[\/latex]\u00a0into the equation for displacement, [latex]x={x}_{0}+\\bar{v}t[\/latex], yielding\r\n<div id=\"m42099-import-auto-id1807031\" class=\"equation\" title=\"Equation 2.44.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]x={x}_{0}+{v}_{0}t+\\frac{1}{2}{\\text{at}}^{2}\\left(\\text{constant}a\\right)\\text{.}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"m42099-fs-id1164906457202\" class=\"example\" title=\"Example 2.10. Calculating Displacement of an Accelerating Object: Dragsters\">\r\n<div class=\"textbox examples\">\r\n<h3><span class=\"cnx-gentext-example cnx-gentext-autogenerated\">Example 3<\/span><span class=\"cnx-gentext-example cnx-gentext-autogenerated\">. <\/span><span class=\"cnx-gentext-example cnx-gentext-t\">Calculating Displacement of an Accelerating Object: Dragsters<\/span><\/h3>\r\n<div class=\"body\">\r\n\r\nDragsters can achieve average accelerations of <span class=\"token\">26.0 m\/s<sup>2<\/sup><\/span>. Suppose such a dragster accelerates from rest at this rate for 5.56 s. How far does it travel in this time?\r\n<div id=\"m42099-import-auto-id2356722\" class=\"figure\" title=\"Figure 2.31.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"275\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101633\/Figure_02_04_02.jpg\" alt=\"Dragster accelerating down a race track.\" width=\"275\" height=\"221\" \/> Figure 6. U.S. Army Top Fuel pilot Tony \u201cThe Sarge\u201d Schumacher begins a race with a controlled burnout. (credit: Lt. Col. William Thurmond. Photo Courtesy of U.S. Army.)[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\r\nDraw a sketch.\r\n<div id=\"m42099-import-auto-id2168040\" class=\"figure\" title=\"Figure 2.32.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"350\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101634\/Figure_02_04_02a.jpg\" alt=\"Acceleration vector arrow pointing toward the right in the positive x direction, labeled a equals twenty-six point 0 meters per second squared. x position graph with initial position at the left end of the graph. The right end of the graph is labeled x equals question mark.\" width=\"350\" height=\"175\" \/> Figure 7.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\"><\/div>\r\n<\/div>\r\nWe are asked to find displacement, which is <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> if we take <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>0<\/sub><\/span> to be zero. (Think about it like the starting line of a race. It can be anywhere, but we call it 0 and measure all other positions relative to it.) We can use the equation [latex]x={x}_{0}+{v}_{0}t+\\frac{1}{2}{{at}}^{2}[\/latex] once we identify <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0<\/sub><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span>, and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> from the statement of the problem.\r\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\r\n1. Identify the knowns. Starting from rest means that <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0\u00a0<\/sub>= 0<\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span> is given as 26.0 m\/s<sup>2<\/sup>\u00a0and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> is given as 5.56 s.\r\n\r\n2. Plug the known values into the equation to solve for the unknown <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>:\r\n<div id=\"m42099-import-auto-id2171407\" class=\"equation\" title=\"Equation 2.45.\">\r\n<div class=\"mediaobject\">\r\n<div id=\"import-auto-id2171407\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]x={x}_{0}+{v}_{0}t+\\frac{1}{2}{\\text{at}}^{2}\\text{.}[\/latex]<\/div>\r\n<div class=\"equation\" style=\"text-align: center;\" data-type=\"equation\"><\/div>\r\n<\/div>\r\n<\/div>\r\nSince the initial position and velocity are both zero, this simplifies to\r\n<div id=\"m42099-import-auto-id2171445\" class=\"equation\" title=\"Equation 2.46.\">\r\n<div class=\"mediaobject\">\r\n<div id=\"import-auto-id2171445\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]x=\\frac{1}{2}{\\text{at}}^{2}\\text{.}[\/latex]<\/div>\r\n<div class=\"equation\" style=\"text-align: center;\" data-type=\"equation\"><\/div>\r\n<\/div>\r\n<\/div>\r\nSubstituting the identified values of <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span> and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> gives\r\n<div id=\"m42099-import-auto-id2169820\" class=\"equation\" title=\"Equation 2.47.\">\r\n<div class=\"mediaobject\">\r\n<div id=\"import-auto-id2169820\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]x=\\frac{1}{2}\\left(\\text{26}\\text{.}{\\text{0 m\/s}}^{2}\\right){\\left(5\\text{.}\\text{56 s}\\right)}^{2}[\/latex],<\/div>\r\n<\/div>\r\n<\/div>\r\nyielding\r\n<div id=\"m42099-import-auto-id2171362\" class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.48.\"><span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x\u00a0<\/em><\/span>= 402 m.<\/span><\/div>\r\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\r\nIf we convert 402 m to miles, we find that the distance covered is very close to one quarter of a mile, the standard distance for drag racing. So the answer is reasonable. This is an impressive displacement in only 5.56 s, but top-notch dragsters can do a quarter mile in even less time than this.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWhat else can we learn by examining the equation [latex]x={x}_{0}+{v}_{0}t+\\frac{1}{2}{\\text{at}}^{2}[\/latex]? We see that:\r\n<div class=\"itemizedlist\">\r\n<ul class=\"itemizedlist\">\r\n \t<li class=\"listitem\">displacement depends on the square of the elapsed time when acceleration is not zero. In Example 3, the dragster covers only one fourth of the total distance in the first half of the elapsed time<\/li>\r\n \t<li class=\"listitem\">if acceleration is zero, then the initial velocity equals average velocity ([latex]{v}_{0}=\\bar{v}[\/latex]) and [latex]x={x}_{0}+{v}_{0}t+\\frac{1}{2}{\\text{at}}^{2}[\/latex] becomes <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x\u00a0<\/em><\/span>=\u00a0<span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>0\u00a0<\/sub>+\u00a0<span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0<\/sub><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"m42099-fs-id1164906443776\" class=\"example\" title=\"Example 2.11. Calculating Final Velocity: Dragsters\">\r\n<div class=\"textbox examples\">\r\n<h3>Example 4: Calculating Final Velocity: Dragsters<\/h3>\r\nCalculate the final velocity of the dragster in Example 3 without using information about time.\r\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\r\nDraw a sketch.\r\n<div id=\"m42099-import-auto-id4179116\" class=\"figure\" title=\"Figure 2.33.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"350\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101641\/Figure_02_04_02b.jpg\" alt=\"Acceleration vector arrow pointing toward the right, labeled twenty-six point zero meters per second squared. Initial velocity equals 0. Final velocity equals question mark.\" width=\"350\" height=\"175\" \/> Figure 8.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\"><\/div>\r\n<\/div>\r\nThe equation [latex]{v}^{2}={v}_{0}^{2}+2a\\left(x-{x}_{0}\\right)[\/latex] is ideally suited to this task because it relates velocities, acceleration, and displacement, and no time information is required.\r\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\r\n1. Identify the known values. We know that <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0\u00a0<\/sub>= 0<\/span>, since the dragster starts from rest. Then we note that <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span>\u2212<span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>0\u00a0<\/sub>= 402 m<\/span> (this was the answer in Example 3). Finally, the average acceleration was given to be <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a\u00a0<\/em><\/span>= 26.0 m\/s<sup>2<\/sup><\/span>.\r\n\r\n2. Plug the knowns into the equation[latex]{v}^{2}={v}_{0}^{2}+2a\\left(x-{x}_{0}\\right)[\/latex] and solve for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span>.<\/span>\r\n<div id=\"m42099-import-auto-id1680164\" class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.51.\"><span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sup>2\u00a0<\/sup>= 0 + 2(26.0 m\/s<sup>2<\/sup>)(402 m).<\/span><\/div>\r\nThus\r\n<div id=\"m42099-import-auto-id2177794\" class=\"equation\" title=\"Equation 2.52.\">\r\n<div class=\"mediaobject\">\r\n<div id=\"import-auto-id2177794\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]{v}^{2}=2\\text{.}\\text{09}\\times {\\text{10}}^{4}{\\text{m}}^{2}{\\text{\/s}}^{2}[\/latex].<\/div>\r\n<div class=\"equation\" style=\"text-align: center;\" data-type=\"equation\"><\/div>\r\n<\/div>\r\n<\/div>\r\nTo get <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span>, we take the square root:\r\n<div id=\"m42099-import-auto-id2177812\" class=\"equation\" title=\"Equation 2.53.\">\r\n<div class=\"mediaobject\">\r\n<div id=\"import-auto-id2177812\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]v=\\sqrt{2\\text{.}\\text{09}\\times {\\text{10}}^{4}{\\text{m}}^{2}{\\text{\/s}}^{2}}=\\text{145 m\/s}[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n<span class=\"bold\"><strong>Discussion<\/strong><\/span>\r\n\r\n145 m\/s is about 522 km\/h or about 324 mi\/h, but even this breakneck speed is short of the record for the quarter mile. Also, note that a square root has two values; we took the positive value to indicate a velocity in the same direction as the acceleration.\r\n\r\n<\/div>\r\n<\/div>\r\nAn examination of the equation [latex]{v}^{2}={v}_{0}^{2}+2a\\left(x-{x}_{0}\\right)[\/latex] can produce further insights into the general relationships among physical quantities:\r\n<div class=\"itemizedlist\">\r\n<ul class=\"itemizedlist\">\r\n \t<li class=\"listitem\">The final velocity depends on how large the acceleration is and the distance over which it acts<\/li>\r\n \t<li class=\"listitem\">For a fixed deceleration, a car that is going twice as fast doesn\u2019t simply stop in twice the distance\u2014it takes much further to stop. (This is why we have reduced speed zones near schools.)<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 id=\"m42099-fs-id1164906446591\"><span class=\"cnx-gentext-section cnx-gentext-t\">Putting Equations Together<\/span><\/h2>\r\nIn the following examples, we further explore one-dimensional motion, but in situations requiring slightly more algebraic manipulation. The examples also give insight into problem-solving techniques. The box below provides easy reference to the equations needed.\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\"><strong><span class=\"cnx-gentext-tip-t\">Summary of Kinematic Equations (constant <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span>)<\/span><\/strong><\/h3>\r\n<div class=\"body\">\r\n<div id=\"m42099-import-auto-id1771742\" class=\"equation\" title=\"Equation 2.54.\">\r\n<div class=\"mediaobject\">\r\n<div id=\"import-auto-id1771742\" class=\"equation\" data-type=\"equation\">\r\n<div id=\"import-auto-id1771742\" class=\"equation\" data-type=\"equation\">[latex]x={x}_{0}+\\bar{v}t[\/latex]<\/div>\r\n<div id=\"import-auto-id4178996\" class=\"equation\" data-type=\"equation\">[latex]\\bar{v}=\\frac{{v}_{0}+v}{2}[\/latex]<\/div>\r\n<div id=\"import-auto-id1680040\" class=\"equation\" data-type=\"equation\">[latex]v={v}_{0}+\\text{at}[\/latex]<\/div>\r\n<div id=\"import-auto-id1680037\" class=\"equation\" data-type=\"equation\">[latex]x={x}_{0}+{v}_{0}t+\\frac{1}{2}{\\text{at}}^{2}[\/latex]<\/div>\r\n<div id=\"import-auto-id2178979\" class=\"equation\" data-type=\"equation\">[latex]{v}^{2}={v}_{0}^{2}+2a\\left(x-{x}_{0}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"m42099-fs-id1164906424692\" class=\"example\" title=\"Example 2.12. Calculating Displacement: How Far Does a Car Go When Coming to a Halt?\">\r\n<div class=\"textbox examples\">\r\n<h3>Example 5. Calculating Displacement: How Far Does a Car Go When Coming to a Halt?<\/h3>\r\nOn dry concrete, a car can decelerate at a rate of <span class=\"token\">7.00 m\/s<sup>2<\/sup><\/span>, whereas on wet concrete it can decelerate at only <span class=\"token\">5.00 m\/s<sup>2<\/sup><\/span>. Find the distances necessary to stop a car moving at 30.0 m\/s (about 110 km\/h) (a) on dry concrete and (b) on wet concrete. (c) Repeat both calculations, finding the displacement from the point where the driver sees a traffic light turn red, taking into account his reaction time of 0.500 s to get his foot on the brake.\r\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\r\nDraw a sketch.\r\n<div id=\"m42099-import-auto-id2174020\" class=\"figure\" title=\"Figure 2.34.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"350\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101649\/Figure_02_04_02c.jpg\" alt=\"Initial velocity equals thirty meters per second. Final velocity equals 0. Acceleration dry equals negative 7 point zero zero meters per second squared. Acceleration wet equals negative 5 point zero zero meters per second squared.\" width=\"350\" height=\"253\" \/> Figure 9.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\"><\/div>\r\n<\/div>\r\nIn order to determine which equations are best to use, we need to list all of the known values and identify exactly what we need to solve for. We shall do this explicitly in the next several examples, using tables to set them off.\r\n<h4><span class=\"bold\"><strong>Solution for (a)<\/strong><\/span><\/h4>\r\n1. Identify the knowns and what we want to solve for. We know that <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0\u00a0<\/sub>= 30.0 m\/s<\/span>; <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v\u00a0<\/em><\/span>= 0<\/span>; a = -7.00 m\/s<sup>2<\/sup>\u00a0(<span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span> is negative because it is in a direction opposite to velocity). We take <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>0<\/sub><\/span> to be 0. We are looking for displacement <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>, or <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span>\u2212<span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>0<\/sub><\/span>.\r\n\r\n2. Identify the equation that will help up solve the problem. The best equation to use is\r\n<div id=\"m42099-import-auto-id2180580\" class=\"equation\" title=\"Equation 2.59.\">\r\n<div class=\"mediaobject\">\r\n<div id=\"import-auto-id2180580\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]{v}^{2}={v}_{0}^{2}+2a\\left(x-{x}_{0}\\right)[\/latex].<\/div>\r\n<div class=\"equation\" style=\"text-align: center;\" data-type=\"equation\"><\/div>\r\n<\/div>\r\n<\/div>\r\nThis equation is best because it includes only one unknown, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>. We know the values of all the other variables in this equation. (There are other equations that would allow us to solve for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>, but they require us to know the stopping time, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span>, which we do not know. We could use them but it would entail additional calculations.)\r\n\r\n3. Rearrange the equation to solve for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>.\r\n<div id=\"m42099-import-auto-id4179294\" class=\"equation\" title=\"Equation 2.60.\">\r\n<div class=\"mediaobject\">\r\n<div id=\"import-auto-id4179294\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]x-{x}_{0}=\\frac{{v}^{2}-{v}_{0}^{2}}{2a}[\/latex]<\/div>\r\n<div class=\"equation\" style=\"text-align: center;\" data-type=\"equation\"><\/div>\r\n<\/div>\r\n<\/div>\r\n4. Enter known values.\r\n<div id=\"m42099-import-auto-id2293184\" class=\"equation\" title=\"Equation 2.61.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]x - 0=\\frac{{0}^{2}-{\\left(\\text{30}\\text{.}\\text{0 m\/s}\\right)}^{2}}{2\\left(-7\\text{.}{\\text{00 m\/s}}^{2}\\right)}[\/latex]<\/div>\r\n<div class=\"mediaobject\"><\/div>\r\n<\/div>\r\nThus,\r\n<div id=\"m42099-import-auto-id2293155\" class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.62.\"><span class=\"token\"> <span class=\"emphasis mathml-mi\"><em>x<\/em><\/span> = 64.3 m on dry concrete.<\/span><\/div>\r\n<h4><span class=\"bold\"><strong>Solution for (b)<\/strong><\/span><\/h4>\r\nThis part can be solved in exactly the same manner as Part A. The only difference is that the deceleration is <span class=\"token\">\u20135.00 m\/s<sup>2<\/sup><\/span>. The result is\r\n<div id=\"m42099-import-auto-id2178635\" class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.63.\"><span class=\"token\"> <span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub> wet <\/sub> = 90.0 m on wet concrete.<\/span><\/div>\r\n<h4><span class=\"bold\"><strong>Solution for (c)<\/strong><\/span><\/h4>\r\nOnce the driver reacts, the stopping distance is the same as it is in Parts A and B for dry and wet concrete. So to answer this question, we need to calculate how far the car travels during the reaction time, and then add that to the stopping time. It is reasonable to assume that the velocity remains constant during the driver\u2019s reaction time.\r\n\r\n1. Identify the knowns and what we want to solve for. We know that [latex]\\bar{v}=30.0 \\text{ m\/s}[\/latex]; <em>t<\/em><sub>reaction<\/sub> = 0.500 s; <em>a<\/em><sub>reaction<\/sub> = 0. We take<i>\u00a0x<sub>0-<\/sub><\/i><sub>reaction\u00a0<\/sub><em>=\u00a0<\/em>to be 0. We are looking for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>reaction<\/sub><\/span>.\r\n\r\n2. Identify the best equation to use.\u00a0[latex]x={x}_{0}+\\bar{v}t[\/latex]\u00a0works well because the only unknown value is <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>, which is what we want to solve for.\r\n\r\n3. Plug in the knowns to solve the equation.\r\n<div id=\"m42099-import-auto-id2175306\" class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.64.\"><span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x\u00a0<\/em><\/span>= 0+(30.0 m\/s)(0.500 s)=15.0 m.<\/span><\/div>\r\n<div class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.64.\"><\/div>\r\nThis means the car travels 15.0 m while the driver reacts, making the total displacements in the two cases of dry and wet concrete 15.0 m greater than if he reacted instantly.\r\n\r\n4. Add the displacement during the reaction time to the displacement when braking.\r\n<div id=\"m42099-import-auto-id1658817\" class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.65.\"><span class=\"token\"> <span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub> braking <\/sub> + <span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub> reaction <\/sub> = <span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub> total<\/sub><\/span><\/div>\r\n<div class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.65.\"><\/div>\r\n<div class=\"orderedlist\">\r\n<ol class=\"orderedlist\" type=\"a\">\r\n \t<li class=\"listitem\">64.3 m + 15.0 m = 79.3 m when dry<\/li>\r\n \t<li class=\"listitem\">90.0 m + 15.0 m = 105 m when wet<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"m42099-import-auto-id1658840\" class=\"figure\" title=\"Figure 2.35.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"600\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101654\/Figure_02_04_03.jpg\" alt=\"Diagram showing the various braking distances necessary for stopping a car. With no reaction time considered, braking distance is 64 point 3 meters on a dry surface and 90 meters on a wet surface. With reaction time of 0 point 500 seconds, braking distance is 79 point 3 meters on a dry surface and 105 meters on a wet surface.\" width=\"600\" height=\"379\" \/> Figure 10. The distance necessary to stop a car varies greatly, depending on road conditions and driver reaction time. Shown here are the braking distances for dry and wet pavement, as calculated in this example, for a car initially traveling at 30.0 m\/s. Also shown are the total distances traveled from the point where the driver first sees a light turn red, assuming a 0.500 s reaction time.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\r\nThe displacements found in this example seem reasonable for stopping a fast-moving car. It should take longer to stop a car on wet rather than dry pavement. It is interesting that reaction time adds significantly to the displacements. But more important is the general approach to solving problems. We identify the knowns and the quantities to be determined and then find an appropriate equation. There is often more than one way to solve a problem. The various parts of this example can in fact be solved by other methods, but the solutions presented above are the shortest.\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 6. Calculating Time: A Car Merges into Traffic<\/h3>\r\n<div class=\"body\">\r\n\r\nSuppose a car merges into freeway traffic on a 200-m-long ramp. If its initial velocity is 10.0 m\/s and it accelerates at <span class=\"token\">2.00 m\/s<sup>2<\/sup><\/span>, how long does it take to travel the 200 m up the ramp? (Such information might be useful to a traffic engineer.)\r\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\r\nDraw a sketch.\r\n<div id=\"m42099-import-auto-id2296882\" class=\"figure\" title=\"Figure 2.36.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"250\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101656\/Figure_02_04_03a.jpg\" alt=\"A line segment with ends labeled x subs zero equals zero and x = two hundred. Above the line segment, the equation t equals question mark indicates that time is unknown. Three vectors, all pointing in the direction of x equals 200, represent the other knowns and unknowns. They are labeled v sub zero equals ten point zero meters per second, v equals question mark, and a equals two point zero zero meters per second squared.\" width=\"250\" height=\"289\" \/> Figure 11.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\"><\/div>\r\n<\/div>\r\nWe are asked to solve for the time <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span>. As before, we identify the known quantities in order to choose a convenient physical relationship (that is, an equation with one unknown, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span>).\r\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\r\n1. Identify the knowns and what we want to solve for. We know that <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0\u00a0<\/sub>= 10 m\/s<\/span>; <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a\u00a0<\/em><\/span>= 2.00 m\/s<sup>2<\/sup><\/span>; and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x\u00a0<\/em><\/span>= 200 m<\/span>.\r\n\r\n2. We need to solve for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span>. Choose the best equation. [latex]x={x}_{0}+{v}_{0}t+\\frac{1}{2}{{at}}^{2}[\/latex]\u00a0works best because the only unknown in the equation is the variable <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> for which we need to solve.\r\n\r\n3. We will need to rearrange the equation to solve for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span>. In this case, it will be easier to plug in the knowns first.\r\n<p style=\"text-align: center;\">[latex]\\text{200 m}=\\text{0 m}+\\left(\\text{10}\\text{.}\\text{0 m\/s}\\right)t+\\frac{1}{2}\\left(2\\text{.}{\\text{00 m\/s}}^{2}\\right){t}^{2}[\/latex]<\/p>\r\n4. Simplify the equation. The units of meters (m) cancel because they are in each term. We can get the units of seconds (s) to cancel by taking <em>t<\/em> = <em>ts<\/em>, where <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> is the magnitude of time and s is the unit. Doing so leaves\r\n<div id=\"m42099-eip-635\" class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.67.\"><span class=\"token\">200 = 10<span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>+\u00a0<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sup>2<\/sup>.<\/span><\/div>\r\n<div class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.67.\"><\/div>\r\n5. Use the quadratic formula to solve for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span><span class=\"emphasis\"><em>. <\/em><\/span>\r\n\r\n(a) Rearrange the equation to get 0 on one side of the equation.\r\n<div id=\"m42099-import-auto-id1680208\" class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.68.\"><span class=\"token\"> <span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sup> 2 <\/sup> + 10<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span> \u2212 200 = 0 <\/span><\/div>\r\nThis is a quadratic equation of the form\r\n<p style=\"text-align: center;\">[latex]{\\text{at}}^{2}+\\text{bt}+c=0[\/latex]<\/p>\r\nwhere the constants are <em>a<\/em> = 1.00, <em>b<\/em> = 10.0 and <em>c<\/em> = -200.\r\n\r\n(b) Its solutions are given by the quadratic formula:\r\n<div id=\"import-auto-id2367246\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]t=\\frac{-b\\pm \\sqrt{{b}^{2}-4\\text{ac}}}{2a}[\/latex]<\/div>\r\n<div class=\"equation\" data-type=\"equation\"><\/div>\r\nThis yields two solutions for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span>, which are\r\n<p style=\"text-align: center;\"><em>t<\/em> = 10.0 and -20.0.<\/p>\r\nIn this case, then, the time is <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>=\u00a0<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> in seconds, or\r\n<p style=\"text-align: center;\"><em>t<\/em> = 10.0 s and -20.0 s.<\/p>\r\nA negative value for time is unreasonable, since it would mean that the event happened 20 s before the motion began. We can discard that solution. Thus,\r\n<p style=\"text-align: center;\"><em>t<\/em> = 10.0 s.<\/p>\r\n\r\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\r\nWhenever an equation contains an unknown squared, there will be two solutions. In some problems both solutions are meaningful, but in others, such as the above, only one solution is reasonable. The 10.0 s answer seems reasonable for a typical freeway on-ramp.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWith the basics of kinematics established, we can go on to many other interesting examples and applications. In the process of developing kinematics, we have also glimpsed a general approach to problem solving that produces both correct answers and insights into physical relationships. <a class=\"link\" title=\"2.6. Problem-Solving Basics for One-Dimensional Kinematics\" href=\".\/chapter\/2-6-problem-solving-basics-for-one-dimensional-kinematics\/\" target=\"_blank\">Problem-Solving Basics<\/a> discusses problem-solving basics and outlines an approach that will help you succeed in this invaluable task.\r\n<div id=\"m42099-fs-id1164906508057\" class=\"note\">\r\n<div class=\"textbox learning-objectives\">\r\n<h3><strong><span class=\"cnx-gentext-tip-t\">Making Connections: Take-Home Experiment\u2014Breaking News<\/span><\/strong><\/h3>\r\n<div class=\"body\">We have been using SI units of meters per second squared to describe some examples of acceleration or deceleration of cars, runners, and trains. To achieve a better feel for these numbers, one can measure the braking deceleration of a car doing a slow (and safe) stop. Recall that, for average acceleration, [latex]\\bar{a}=\\Delta v\/ \\Delta t[\/latex]. While traveling in a car, slowly apply the brakes as you come up to a stop sign. Have a passenger note the initial speed in miles per hour and the time taken (in seconds) to stop. From this, calculate the deceleration in miles per hour per second. Convert this to meters per second squared and compare with other decelerations mentioned in this chapter. Calculate the distance traveled in braking.<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Check Your Understanding<\/h3>\r\n<div id=\"m42099-fs-id1164906508057\" class=\"note\">\r\n\r\nA manned rocket accelerates at a rate of <span class=\"token\">20 m\/s<sup>2<\/sup><\/span> during launch. How long does it take the rocket reach a velocity of 400 m\/s?\r\n\r\n<\/div>\r\n<div id=\"m42099-fs-id1164906458555\" class=\"solution\">\r\n<h4 class=\"title\"><strong><span class=\"epub-only pre-text\">Solution<\/span><\/strong><\/h4>\r\n<div class=\"body\">\r\n\r\nTo answer this, choose an equation that allows you to solve for time <span class=\"emphasis\"><em><span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span><\/em><\/span>, given only <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0<\/sub><\/span>, and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span>.\r\n<p style=\"text-align: center;\">[latex]v={v}_{0}+{at}[\/latex]<\/p>\r\nRearrange to solve for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span><span class=\"emphasis\"><em>. <\/em><\/span>\r\n<div id=\"m42099-import-auto-id2168930\" class=\"equation\" title=\"Equation 2.75.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]t=\\frac{v-v{}_{0}\\text{}}{a}=\\frac{\\text{400 m\/s}-\\text{0 m\/s}}{{\\text{20 m\/s}}^{2}}=\\text{20 s}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h2>Section Summary<\/h2>\r\n<ul>\r\n \t<li>To simplify calculations we take acceleration to be constant, so that [latex]\\bar{a}=a[\/latex] at all times.<\/li>\r\n \t<li>We also take initial time to be zero.<\/li>\r\n \t<li><span style=\"color: #000000;\">Initial position and velocity are given a subscript 0; final values have no subscript. Thus,<\/span>\r\n[latex]\\begin{cases}{\\Delta}{t} &amp;=&amp; t \\\\{\\Delta}{x} &amp;=&amp; x-{{x}_{0}}\\\\{\\Delta}{v} &amp;=&amp; v-{{v}_{0}}\\end{cases}[\/latex]<\/li>\r\n \t<li>The following kinematic equations for motion with constant [latex]a[\/latex] are useful:\r\n<div id=\"import-auto-id1771742\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]x={x}_{0}+\\bar{v}t[\/latex]<\/div>\r\n<div id=\"import-auto-id4178996\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\bar{v}=\\frac{{v}_{0}+v}{2}[\/latex]<\/div>\r\n<div id=\"import-auto-id1680040\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]v={v}_{0}+\\text{at}[\/latex]<\/div>\r\n<div id=\"import-auto-id1680037\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]x={x}_{0}+{v}_{0}t+\\frac{1}{2}{\\text{at}}^{2}[\/latex]<\/div>\r\n<div id=\"import-auto-id2178979\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]{v}^{2}={v}_{0}^{2}+2a\\left(x-{x}_{0}\\right)[\/latex]<\/div>\r\nIn vertical motion, <em>y<\/em>\u00a0is substituted for <em>x<\/em>.<\/li>\r\n<\/ul>\r\n<div class=\"textbox exercises\">\r\n<h3>Problems &amp; Exercises<\/h3>\r\n1. An Olympic-class sprinter starts a race with an acceleration of 4.50 m\/s<sup>2<\/sup>. (a) What is her speed 2.40 s later? (b) Sketch a graph of her position vs. time for this period.\r\n\r\n2. A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10 \u00d7 10<sup>4<\/sup> m\/s<sup>2<\/sup>, and 1.85 ms (1 ms = 10-3 s) elapses from the time the ball first touches the mitt until it stops, what was the initial velocity of the ball?\r\n\r\n3. A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 6.20 \u00d7 10<sup>5\u00a0<\/sup>m\/s<sup>2<\/sup> for\u00a08.10 \u00d7 10<sup>-4 <\/sup>s.\u00a0What is its muzzle velocity (that is, its final velocity)?\r\n\r\n4.\u00a0(a) A light-rail commuter train accelerates at a rate of <span class=\"token\">1.35 m\/s<sup>2<\/sup><\/span>. How long does it take to reach its top speed of 80.0 km\/h, starting from rest?\u00a0(b) The same train ordinarily decelerates at a rate of <span class=\"token\">1.65 m\/s<sup>2<\/sup><\/span>. How long does it take to come to a stop from its top speed?\u00a0(c) In emergencies the train can decelerate more rapidly, coming to rest from 80.0 km\/h in 8.30 s. What is its emergency deceleration in <span class=\"token\">m\/s<sup>2<\/sup><\/span>?\r\n\r\n5.\u00a0While entering a freeway, a car accelerates from rest at a rate of <span class=\"token\">2.40 m\/s<sup>2<\/sup><\/span> for 12.0 s.\u00a0(a) Draw a sketch of the situation.\u00a0(b) List the knowns in this problem.\u00a0(c) How far does the car travel in those 12.0 s?\u00a0To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, check your units, and discuss whether the answer is reasonable.\u00a0(d) What is the car\u2019s final velocity? Solve for this unknown in the same manner as in part (c), showing all steps explicitly.\r\n\r\n6. At the end of a race, a runner decelerates from a velocity of 9.00 m\/s at a rate of <span class=\"token\">2.00 m\/s<sup>2<\/sup><\/span>.\u00a0(a) How far does she travel in the next 5.00 s?\u00a0(b) What is her final velocity?\u00a0(c) Evaluate the result. Does it make sense?\r\n\r\n7. <strong>Professional Application:\u00a0<\/strong>Blood is accelerated from rest to 30.0 cm\/s in a distance of 1.80 cm by the left ventricle of the heart. (a) Make a sketch of the situation. (b) List the knowns in this problem. (c) How long does the acceleration take? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units. (d) Is the answer reasonable when compared with the time for a heartbeat?\r\n\r\n8. In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m\/s to 40.0 m\/s in the same direction. If this shot takes 3.33 \u00d7 10<sup>-2<\/sup>, calculate the distance over which the puck accelerates.\r\n\r\n9. A powerful motorcycle can accelerate from rest to 26.8 m\/s (100 km\/h) in only 3.90 s. (a) What is its average acceleration? (b) How far does it travel in that time?\r\n\r\n10. Freight trains can produce only relatively small accelerations and decelerations. (a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m\/s<sup>2<\/sup> for 8.00 min, starting with an initial velocity of 4.00 m\/s? (b) If the train can slow down at a rate of 0.550 m\/s<sup>2<\/sup>, how long will it take to come to a stop from this velocity? (c) How far will it travel in each case?\r\n\r\n11.\u00a0A fireworks shell is accelerated from rest to a velocity of 65.0 m\/s over a distance of 0.250 m. (a) How long did the acceleration last? (b) Calculate the acceleration.\r\n\r\n12. A swan on a lake gets airborne by flapping its wings and running on top of the water. (a) If the swan must reach a velocity of 6.00 m\/s to take off and it accelerates from rest at an average rate of 0.350 m\/s<sup>2<\/sup>, how far will it travel before becoming airborne? (b) How long does this take?\r\n\r\n13. <strong>Professional Application:\u00a0<\/strong>A woodpecker\u2019s brain is specially protected from large decelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker\u2019s head comes to a stop from an initial velocity of 0.600 m\/s in a distance of only 2.00 mm. (a) Find the acceleration in m\/s<sup>2<\/sup> and in multiples of <em>g<\/em> (<em>g<\/em> = 9.80 m\/s<sup>2<\/sup>. (b) Calculate the stopping time. (c) The tendons cradling the brain stretch, making its stopping distance 4.50 mm (greater than the head and, hence, less deceleration of the brain). What is the brain\u2019s deceleration, expressed in multiples of\u00a0<em>g<\/em>?\r\n\r\n14. An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m\/s and comes to a full stop after compressing the padding and his body 0.350 m. (a) What is his deceleration? (b) How long does the collision last?\r\n\r\n15. In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no parachute to escape certain death. Some fell about 20,000 feet (6000 m), and some of them survived, with few life-threatening injuries. For these lucky pilots, the tree branches and snow drifts on the ground allowed their deceleration to be relatively small. If we assume that a pilot\u2019s speed upon impact was 123 mph (54 m\/s), then what was his deceleration? Assume that the trees and snow stopped him over a distance of 3.0 m.\r\n\r\n16. Consider a grey squirrel falling out of a tree to the ground. (a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel\u2019s velocity just before hitting the ground, assuming it fell from a height of 3.0 m. (b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem.\r\n\r\n17. An express train passes through a station. It enters with an initial velocity of 22.0 m\/s and decelerates at a rate of [latex]0text{.}{text{150 m\/s}}^{2}[\/latex] as it goes through. The station is 210 m long. (a) How long is the nose of the train in the station? (b) How fast is it going when the nose leaves the station? (c) If the train is 130 m long, when does the end of the train leave the station? (d) What is the velocity of the end of the train as it leaves?\r\n\r\n18. Dragsters can actually reach a top speed of 145 m\/s in only 4.45 s\u2014considerably less time than given in Example 2.10 and Example 2.11. (a) Calculate the average acceleration for such a dragster. (b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any information on time. (c) Why is the final velocity greater than that used to find the average acceleration? Hint: Consider whether the assumption of constant acceleration is valid for a dragster. If not, discuss whether the acceleration would be greater at the beginning or end of the run and what effect that would have on the final velocity.\r\n\r\n19. A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m\/s and accelerates at the rate of 0.500 m\/s<sup>2<\/sup>\u00a0for 7.00 s. (a) What is his final velocity? (b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save? (c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m\/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?\r\n\r\n20. In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, with a maximum speed of 183.58 mi\/h. The one-way course was 5.00 mi long. Acceleration rates are often described by the time it takes to reach 60.0 mi\/h from rest. If this time was 4.00 s, and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course?\r\n\r\n21. (a) A world record was set for the men\u2019s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt \"coasted\" across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration. (b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Selected Solutions to Problems &amp; Exercises<\/h3>\r\n1.\u00a010.8 m\/s\r\n\r\n(b)<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1322\/2015\/12\/03205527\/unnumbered_art_p44.jpg\" alt=\"Line graph of position in meters versus time in seconds. The line begins at the origin and is concave up, with its slope increasing over time.\" width=\"350\" height=\"441\" data-media-type=\"image\/jpg\" \/>\r\n\r\n2.\u00a038.9 m\/s (about 87 miles per hour)\r\n\r\n4. (a) 16.5 s (b) 13.5 s (c) -2.68 m\/s<sup>2<\/sup>\r\n\r\n6.\u00a0(a)\u00a020.0 m (b) -1.00 m\/s\u00a0(c) This result does not really make sense. If the runner starts at 9.00 m\/s and decelerates at 2.00 m\/s<sup>2<\/sup>, then she will have stopped after 4.50 s. If she continues to decelerate, she will be running backwards.\r\n\r\n8. 0.799 m\r\n\r\n10. (a) 28.0 m\/s (b) 50.9 s\u00a0(c) 7.68 km to accelerate and 713 m to decelerate\r\n\r\n12. (a) 51.4 m (b) 17.1 s\r\n\r\n14. (a) -80\u00a0m\/s<sup>2\u00a0<\/sup>(b)\u00a09.33 \u00d7 10<sup>-<\/sup><sup>2<\/sup> s\r\n\r\n16. (a) 7.7 m\/s (b) -15 \u00d7 10<sup>2<\/sup>\u00a0m\/s<sup>2\u00a0<\/sup>his is about 3 times the deceleration of the pilots, who were falling from thousands of meters high!\r\n\r\n18.\u00a0(a)\u00a036.2\u00a0m\/s<sup>2\u00a0<\/sup>(b)\u00a0162 m\/s\u00a0(c) v &gt; v<sub>max<\/sub>, because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears, and would have a greater acceleration in first gear than second gear than third gear, etc. The acceleration would be greatest at the beginning, so it would not be accelerating at 32 m\/s<sup>2<\/sup> during the last few meters, but substantially less, and the final velocity would be less than 162 m\/s.\r\n\r\n20.\u00a0104 s\r\n\r\n21.\u00a0(a)\u00a0<em>v<\/em> = 12\/2 m\/s; <em>a<\/em> = 4.07 m\/s<sup>2\u00a0<\/sup>(b)<em> v<\/em> = 11.2 m\/s\r\n\r\n<\/div>\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<div class=\"itemizedlist\">\n<ul class=\"itemizedlist\">\n<li class=\"listitem\">Calculate displacement of an object that is not accelerating, given initial position and velocity.<\/li>\n<li class=\"listitem\">Calculate final velocity of an accelerating object, given initial velocity, acceleration, and time.<\/li>\n<li class=\"listitem\">Calculate displacement and final position of an accelerating object, given initial position, initial velocity, time, and acceleration.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"m42099-import-auto-id1489960\" class=\"figure\" title=\"Figure 2.25.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 310px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101608\/Figure_02_05_00.jpg\" alt=\"Four men racing up a river in their kayaks.\" width=\"300\" height=\"404\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. Kinematic equations can help us describe and predict the motion of moving objects such as these kayaks racing in Newbury, England. (credit: Barry Skeates, Flickr)<\/p>\n<\/div>\n<p>We might know that the greater the acceleration of, say, a car moving away from a stop sign, the greater the displacement in a given time. But we have not developed a specific equation that relates acceleration and displacement. In this section, we develop some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration already covered.<\/p>\n<\/div>\n<\/div>\n<h2><span class=\"cnx-gentext-section cnx-gentext-t\">Notation: <span class=\"emphasis\"><em>t<\/em><\/span>, <span class=\"emphasis\"><em>x<\/em><\/span>, <span class=\"emphasis\"><em>v<\/em><\/span>, <span class=\"emphasis\"><em>a<\/em><\/span><\/span><\/h2>\n<p>First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. Since elapsed time is <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>=\u00a0<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>f<\/sub>\u2212<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>0<\/sub><\/span>, taking <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>0\u00a0<\/sub>= 0<\/span> means that <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>=\u00a0<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>f<\/sub><\/span>, the final time on the stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity. That is, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>0<\/sub><\/span> <span class=\"emphasis\"><em>is the initial position<\/em><\/span> and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0<\/sub><\/span> <span class=\"emphasis\"><em>is the initial velocity<\/em><\/span>. We put no subscripts on the final values. That is, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> <span class=\"emphasis\"><em> is the final time<\/em><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> <span class=\"emphasis\"><em> is the final position<\/em><\/span>, and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span> <span class=\"emphasis\"><em>is the final velocity<\/em><\/span>. This gives a simpler expression for elapsed time\u2014now, <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span>=<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span>. It also simplifies the expression for displacement, which is now <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>x\u00a0<\/em><\/span>=\u00a0<span class=\"emphasis mathml-mi\"><em>x<\/em><\/span>\u2212<span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>0<\/sub><\/span>. Also, it simplifies the expression for change in velocity, which is now <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>v\u00a0<\/em><\/span>=\u00a0<span class=\"emphasis mathml-mi\"><em>v<\/em><\/span>\u2212<span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0<\/sub><\/span>. <span style=\"color: #000000;\">To summarize, using the simplified notation, with the initial time taken to be zero,<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{cases}{\\Delta}{t} &=& t \\\\{\\Delta}{x} &=& x-{{x}_{0}}\\\\{\\Delta}{v} &=& v-{{v}_{0}}\\end{cases}[\/latex]<\/p>\n<p>where <span class=\"emphasis\"><em>the subscript 0 denotes an initial value and the absence of a subscript denotes a final value<\/em><\/span> in whatever motion is under consideration.<\/p>\n<p>We now make the important assumption that <span class=\"emphasis\"><em>acceleration is constant<\/em><\/span>. This assumption allows us to avoid using calculus to find instantaneous acceleration. Since acceleration is constant, the average and instantaneous accelerations are equal. That is,<\/p>\n<div id=\"m42099-import-auto-id1819116\" class=\"equation\" title=\"Equation 2.29.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]\\bar{a }=a=\\text{ constant}[\/latex],<\/div>\n<\/div>\n<p>so we use the symbol <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span> for acceleration at all times. Assuming acceleration to be constant does not seriously limit the situations we can study nor degrade the accuracy of our treatment. For one thing, acceleration <span class=\"emphasis\"><em>is<\/em><\/span> constant in a great number of situations. Furthermore, in many other situations we can accurately describe motion by assuming a constant acceleration equal to the average acceleration for that motion. Finally, in motions where acceleration changes drastically, such as a car accelerating to top speed and then braking to a stop, the motion can be considered in separate parts, each of which has its own constant acceleration.<\/p>\n<div class=\"section textbox shaded\" title=\"Notation: t, x, v, a\">\n<h3><strong><span class=\"cnx-gentext-tip-t\">Solving for Displacement (<span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>) and Final Position (<span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>) from Average Velocity when Acceleration (<span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span>) is Constant<\/span><\/strong><\/h3>\n<div id=\"m42099-fs-id1164906424651\" class=\"note\">\n<div class=\"body\">\n<p>To get our first two new equations, we start with the definition of average velocity:<\/p>\n<div id=\"m42099-import-auto-id1782525\" class=\"equation\" title=\"Equation 2.30.\">\n<div class=\"mediaobject\">\n<div id=\"import-auto-id1782525\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\bar{v}=\\frac{\\Delta x}{\\Delta t}[\/latex].<\/div>\n<div class=\"equation\" style=\"text-align: center;\" data-type=\"equation\"><\/div>\n<\/div>\n<\/div>\n<p>Substituting the simplified notation for <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> and <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> yields<\/p>\n<div id=\"m42099-import-auto-id4145482\" class=\"equation\" title=\"Equation 2.31.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]\\bar{v}=\\frac{x-{x}_{0}}{t}[\/latex]<\/div>\n<div class=\"mediaobject\" style=\"text-align: center;\"><\/div>\n<\/div>\n<p>Solving for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> yields<\/p>\n<div id=\"m42099-import-auto-id2297251\" class=\"equation\" title=\"Equation 2.32.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]x={x}_{0}+\\bar{v}t[\/latex],<\/div>\n<div class=\"mediaobject\" style=\"text-align: center;\"><\/div>\n<\/div>\n<p>where the average velocity is<\/p>\n<div class=\"MathJax_Display\" style=\"text-align: center;\">[latex]\\bar{v}=\\frac{{v}_{0}+v}{2}\\left(\\text{ constant }a\\right)[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"section\" title=\"Putting Equations Together\">\n<div class=\"titlepage\">\n<p>The equation [latex]\\bar{v}=\\frac{{v}_{0}+v}{2}[\/latex] reflects the fact that, when acceleration is constant, <em>v<\/em> is just the simple average of the initial and final velocities. For example, if you steadily increase your velocity (that is, with constant acceleration) from 30 to 60 km\/h, then your average velocity during this steady increase is 45 km\/h. Using the equation [latex]\\bar{v}=\\frac{{v}_{0}+v}{2}[\/latex]\u00a0to check this, we see that<\/p>\n<div id=\"import-auto-id4152098\" data-type=\"equation\">\n<div class=\"MathJax_Display\" style=\"text-align: center;\">[latex]\\bar{v}=\\frac{{v}_{0}+v}{2}=\\frac{\\text{30 km\/h}+\\text{60 km\/h}}{2}=\\text{45 km\/h}[\/latex],<\/div>\n<\/div>\n<p id=\"import-auto-id2297106\">which seems logical.<\/p>\n<div id=\"m42099-fs-id1164906442368\" class=\"example\" title=\"Example 2.8. Calculating Displacement: How Far does the Jogger Run?\">\n<div class=\"textbox examples\">\n<h3>Example 1. Calculating Displacement: How Far does the Jogger Run?<\/h3>\n<p>A jogger runs down a straight stretch of road with an average velocity of 4.00 m\/s for 2.00 min. What is his final position, taking his initial position to be zero?<\/p>\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\n<p>Draw a sketch.<\/p>\n<div id=\"m42099-import-auto-id2297327\" class=\"figure\" title=\"Figure 2.26.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 410px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101613\/Figure_02_05_00a.jpg\" alt=\"Velocity vector arrow labeled v equals 4 point zero zero meters per second over an x axis displaying initial and final positions. Final position is labeled x equals question mark.\" width=\"400\" height=\"175\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"title\"><\/div>\n<\/div>\n<p>The final position <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> is given by the equation<\/p>\n<div id=\"m42099-import-auto-id2168543\" class=\"equation\" title=\"Equation 2.35.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]x={x}_{0}+\\bar{v}t[\/latex].<\/div>\n<div class=\"mediaobject\" style=\"text-align: center;\"><\/div>\n<\/div>\n<p>To find <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>, we identify the values of <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>0<\/sub><\/span>,\u00a0[latex]\\bar{v}[\/latex], and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> from the statement of the problem and substitute them into the equation.<\/p>\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\n<div id=\"m42099-import-auto-id2300752\" class=\"equation\" title=\"Equation 2.36.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">\n<p id=\"import-auto-id2169540\" style=\"text-align: left;\">1. Identify the knowns. [latex]\\bar{v}=4.00\\text{ m\/s}[\/latex], [latex]\\Delta t=2.00\\text{ min}[\/latex], and [latex]{x}_{0}=0\\text{ m}[\/latex].<\/p>\n<p id=\"import-auto-id2300750\" style=\"text-align: left;\">2. Enter the known values into the equation.<\/p>\n<div id=\"import-auto-id2300752\" class=\"equation\" data-type=\"equation\">[latex]x={x}_{0}+\\bar{v}t=0+\\left(4\\text{.}\\text{00 m\/s}\\right)\\left(\\text{120 s}\\right)=\\text{480 m}[\/latex]<\/div>\n<\/div>\n<\/div>\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\n<p>Velocity and final displacement are both positive, which means they are in the same direction.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>The equation [latex]x={x}_{0}+\\bar{v}t[\/latex]\u00a0gives insight into the relationship between displacement, average velocity, and time. It shows, for example, that displacement is a linear function of average velocity. (By linear function, we mean that displacement depends on\u00a0[latex]\\bar{v}[\/latex]<em data-effect=\"italics\">,<\/em>\u00a0rather than on\u00a0[latex]\\bar{v}[\/latex]\u00a0raised to some other power, such as\u00a0[latex]\\bar{v}^{2}[\/latex]. When graphed, linear functions look like straight lines with a constant slope.) On a car trip, for example, we will get twice as far in a given time if we average 90 km\/h than if we average 45 km\/h.<\/p>\n<div id=\"m42099-import-auto-id1962019\" class=\"figure\" title=\"Figure 2.27.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 310px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101620\/Figure_02_05_00b.jpg\" alt=\"Line graph showing displacement in meters versus average velocity in meters per second. The line is straight with a positive slope. Displacement x increases linearly with increase in average velocity v.\" width=\"300\" height=\"446\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. There is a linear relationship between displacement and average velocity. For a given time t, an object moving twice as fast as another object will move twice as far as the other object.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"title\"><\/div>\n<\/div>\n<div id=\"m42099-fs-id1164906431223\" class=\"note\">\n<div class=\"title textbox shaded\">\n<h3 class=\"title\"><strong><span class=\"cnx-gentext-tip-t\">Solving for Final Velocity<\/span><\/strong><\/h3>\n<div class=\"body\">\n<p>We can derive another useful equation by manipulating the definition of acceleration.<\/p>\n<div id=\"m42099-import-auto-id2366011\" class=\"equation\" title=\"Equation 2.37.\">\n<div class=\"mediaobject\">\n<div id=\"import-auto-id2366011\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]a=\\frac{\\Delta v}{\\Delta t}[\/latex]<\/div>\n<div class=\"equation\" style=\"text-align: center;\" data-type=\"equation\"><\/div>\n<\/div>\n<\/div>\n<p>Substituting the simplified notation for <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span> and <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> gives us<\/p>\n<div id=\"m42099-import-auto-id1962571\" class=\"equation\" title=\"Equation 2.38.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]a=\\frac{v-{v}_{0}}{t}\\text{}\\left(\\text{constant}a\\right)[\/latex]<\/div>\n<div class=\"mediaobject\"><\/div>\n<\/div>\n<p>Solving for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span> yields<\/p>\n<div id=\"m42099-import-auto-id2168495\" class=\"equation\" title=\"Equation 2.39.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]v={v}_{0}+\\text{at}\\text{}\\left(\\text{constant}a\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"m42099-fs-id1164906431414\" class=\"example\" title=\"Example 2.9. Calculating Final Velocity: An Airplane Slowing Down after Landing\">\n<div class=\"textbox examples\">\n<h3><span class=\"cnx-gentext-example cnx-gentext-autogenerated\">Example <\/span><span class=\"cnx-gentext-example cnx-gentext-n\">2<\/span><span class=\"cnx-gentext-example cnx-gentext-autogenerated\">. <\/span><span class=\"cnx-gentext-example cnx-gentext-t\">Calculating Final Velocity: An Airplane Slowing Down after Landing<\/span><\/h3>\n<p>An airplane lands with an initial velocity of 70.0 m\/s and then decelerates at <span class=\"token\">1.50 m\/s<sup>2<\/sup><\/span> for 40.0 s. What is its final velocity?<\/p>\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\n<p>Draw a sketch. We draw the acceleration vector in the direction opposite the velocity vector because the plane is decelerating.<\/p>\n<div id=\"m42099-import-auto-id2300963\" class=\"figure\" title=\"Figure 2.28.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 360px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101623\/Figure_02_05_00c.jpg\" alt=\"Velocity vector arrow pointing toward the right in the positive x direction. Initial velocity equals seventy meters per second. Final velocity equals question mark. An acceleration vector arrow pointing toward the left labeled a equals negative 1 point 50 meters per second squared.\" width=\"350\" height=\"209\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\n<p>1. Identify the knowns. v<sub>0\u00a0<\/sub><span class=\"token\">= 70.0 m\/s<\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a\u00a0<\/em><\/span>= \u22121.50 m\/s<sup>2<\/sup><\/span>, <em>t<\/em> = 40.0 s.<\/p>\n<p>2. Identify the unknown. In this case, it is final velocity, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>f<\/sub><\/span><sub>.<\/sub><\/p>\n<p>3. Determine which equation to use. We can calculate the final velocity using the equation [latex]v={v}_{0}+{at}[\/latex].<\/p>\n<p>4. Plug in the known values and solve.<\/p>\n<div id=\"m42099-import-auto-id2177855\" class=\"equation\" title=\"Equation 2.40.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]v={v}_{0}+\\text{at}=\\text{70}\\text{.}\\text{0 m\/s}+\\left(-1\\text{.}{\\text{50 m\/s}}^{2}\\right)\\left(\\text{40}\\text{.}\\text{0 s}\\right)=\\text{10}\\text{.}\\text{0 m\/s}[\/latex]<\/div>\n<\/div>\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\n<p>The final velocity is much less than the initial velocity, as desired when slowing down, but still positive. With jet engines, reverse thrust could be maintained long enough to stop the plane and start moving it backward. That would be indicated by a negative final velocity, which is not the case here.<\/p>\n<div id=\"m42099-import-auto-id2173965\" class=\"figure\" title=\"Figure 2.29.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 610px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101626\/Figure_02_04_01.jpg\" alt=\"An airplane moving toward the right at two points in time. At time equals 0 the velocity vector arrow points toward the right and is labeled seventy meters per second. The acceleration vector arrow points toward the left and is labeled negative 1 point 5 meters per second squared. At time equals forty seconds, the velocity arrow is shorter, points toward the right, and is labeled ten meters per second. The acceleration vector arrow is still pointing toward the left and is labeled a equals negative 1 point 5 meters per second squared.\" width=\"600\" height=\"202\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5. The airplane lands with an initial velocity of 70.0 m\/s and slows to a final velocity of 10.0 m\/s before heading for the terminal. Note that the acceleration is negative because its direction is opposite to its velocity, which is positive.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>In addition to being useful in problem solving, the equation [latex]v={v}_{0}+\\text{at}[\/latex]\u00a0gives us insight into the relationships among velocity, acceleration, and time. From it we can see, for example, that<\/p>\n<div class=\"itemizedlist\">\n<ul class=\"itemizedlist\">\n<li class=\"listitem\">final velocity depends on how large the acceleration is and how long it lasts<\/li>\n<li class=\"listitem\">if the acceleration is zero, then the final velocity equals the initial velocity <span class=\"token\">(<span class=\"emphasis mathml-mi\"><em>v\u00a0<\/em><\/span>=\u00a0<span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0<\/sub>)<\/span>, as expected (i.e., velocity is constant)<\/li>\n<li class=\"listitem\">if <span class=\"emphasis\"><em><span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span><\/em><\/span> is negative, then the final velocity is less than the initial velocity<\/li>\n<\/ul>\n<\/div>\n<p>(All of these observations fit our intuition, and it is always useful to examine basic equations in light of our intuition and experiences to check that they do indeed describe nature accurately.)<\/p>\n<div id=\"m42099-fs-id1164906437247\" class=\"note\">\n<div class=\"textbox learning-objectives\">\n<h3><span class=\"cnx-gentext-tip-t\">Making Connections: Real-World Connection<\/span><\/h3>\n<section>\n<div data-type=\"note\">\n<div class=\"body\">\n<p>An intercontinental ballistic missile (ICBM) has a larger average acceleration than the Space Shuttle and achieves a greater velocity in the first minute or two of flight (actual ICBM burn times are classified\u2014short-burn-time missiles are more difficult for an enemy to destroy). But the Space Shuttle obtains a greater final velocity, so that it can orbit the earth rather than come directly back down as an ICBM does. The Space Shuttle does this by accelerating for a longer time.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<div class=\"title textbox shaded\">\n<div id=\"m42099-fs-id1164906444525\" class=\"note\">\n<h3 class=\"title\"><strong><span class=\"cnx-gentext-tip-t\">Solving for Final Position When Velocity is Not Constant (<span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a\u00a0<\/em><\/span>\u2260 0<\/span>)<\/span><\/strong><\/h3>\n<div class=\"body\">\n<p>We can combine the equations above to find a third equation that allows us to calculate the final position of an object experiencing constant acceleration. We start with<\/p>\n<div id=\"m42099-import-auto-id2166975\" class=\"equation\" title=\"Equation 2.41.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]v={v}_{0}+{at}[\/latex]<\/div>\n<div class=\"mediaobject\" style=\"text-align: center;\"><\/div>\n<\/div>\n<p>Adding <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0<\/sub><\/span> to each side of this equation and dividing by 2 gives<\/p>\n<div id=\"m42099-import-auto-id1689620\" class=\"equation\" title=\"Equation 2.42.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]\\frac{{v}_{0}+v}{2}={v}_{0}+\\frac{1}{2}{at}[\/latex]<\/div>\n<div class=\"mediaobject\" style=\"text-align: center;\"><\/div>\n<\/div>\n<p>Since [latex]\\frac{{v}_{0}+v}{2}=\\bar{v}[\/latex]\u00a0for constant acceleration, then<\/p>\n<div id=\"m42099-import-auto-id2301379\" class=\"equation\" title=\"Equation 2.43.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]\\bar{v}={v}_{0}+\\frac{1}{2}{at}[\/latex]<\/div>\n<div class=\"mediaobject\"><\/div>\n<\/div>\n<p>Now we substitute this expression for [latex]\\bar{v}[\/latex]\u00a0into the equation for displacement, [latex]x={x}_{0}+\\bar{v}t[\/latex], yielding<\/p>\n<div id=\"m42099-import-auto-id1807031\" class=\"equation\" title=\"Equation 2.44.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]x={x}_{0}+{v}_{0}t+\\frac{1}{2}{\\text{at}}^{2}\\left(\\text{constant}a\\right)\\text{.}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"m42099-fs-id1164906457202\" class=\"example\" title=\"Example 2.10. Calculating Displacement of an Accelerating Object: Dragsters\">\n<div class=\"textbox examples\">\n<h3><span class=\"cnx-gentext-example cnx-gentext-autogenerated\">Example 3<\/span><span class=\"cnx-gentext-example cnx-gentext-autogenerated\">. <\/span><span class=\"cnx-gentext-example cnx-gentext-t\">Calculating Displacement of an Accelerating Object: Dragsters<\/span><\/h3>\n<div class=\"body\">\n<p>Dragsters can achieve average accelerations of <span class=\"token\">26.0 m\/s<sup>2<\/sup><\/span>. Suppose such a dragster accelerates from rest at this rate for 5.56 s. How far does it travel in this time?<\/p>\n<div id=\"m42099-import-auto-id2356722\" class=\"figure\" title=\"Figure 2.31.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 285px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101633\/Figure_02_04_02.jpg\" alt=\"Dragster accelerating down a race track.\" width=\"275\" height=\"221\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. U.S. Army Top Fuel pilot Tony \u201cThe Sarge\u201d Schumacher begins a race with a controlled burnout. (credit: Lt. Col. William Thurmond. Photo Courtesy of U.S. Army.)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\n<p>Draw a sketch.<\/p>\n<div id=\"m42099-import-auto-id2168040\" class=\"figure\" title=\"Figure 2.32.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 360px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101634\/Figure_02_04_02a.jpg\" alt=\"Acceleration vector arrow pointing toward the right in the positive x direction, labeled a equals twenty-six point 0 meters per second squared. x position graph with initial position at the left end of the graph. The right end of the graph is labeled x equals question mark.\" width=\"350\" height=\"175\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"title\"><\/div>\n<\/div>\n<p>We are asked to find displacement, which is <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> if we take <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>0<\/sub><\/span> to be zero. (Think about it like the starting line of a race. It can be anywhere, but we call it 0 and measure all other positions relative to it.) We can use the equation [latex]x={x}_{0}+{v}_{0}t+\\frac{1}{2}{{at}}^{2}[\/latex] once we identify <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0<\/sub><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span>, and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> from the statement of the problem.<\/p>\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\n<p>1. Identify the knowns. Starting from rest means that <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0\u00a0<\/sub>= 0<\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span> is given as 26.0 m\/s<sup>2<\/sup>\u00a0and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> is given as 5.56 s.<\/p>\n<p>2. Plug the known values into the equation to solve for the unknown <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>:<\/p>\n<div id=\"m42099-import-auto-id2171407\" class=\"equation\" title=\"Equation 2.45.\">\n<div class=\"mediaobject\">\n<div id=\"import-auto-id2171407\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]x={x}_{0}+{v}_{0}t+\\frac{1}{2}{\\text{at}}^{2}\\text{.}[\/latex]<\/div>\n<div class=\"equation\" style=\"text-align: center;\" data-type=\"equation\"><\/div>\n<\/div>\n<\/div>\n<p>Since the initial position and velocity are both zero, this simplifies to<\/p>\n<div id=\"m42099-import-auto-id2171445\" class=\"equation\" title=\"Equation 2.46.\">\n<div class=\"mediaobject\">\n<div id=\"import-auto-id2171445\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]x=\\frac{1}{2}{\\text{at}}^{2}\\text{.}[\/latex]<\/div>\n<div class=\"equation\" style=\"text-align: center;\" data-type=\"equation\"><\/div>\n<\/div>\n<\/div>\n<p>Substituting the identified values of <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span> and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> gives<\/p>\n<div id=\"m42099-import-auto-id2169820\" class=\"equation\" title=\"Equation 2.47.\">\n<div class=\"mediaobject\">\n<div id=\"import-auto-id2169820\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]x=\\frac{1}{2}\\left(\\text{26}\\text{.}{\\text{0 m\/s}}^{2}\\right){\\left(5\\text{.}\\text{56 s}\\right)}^{2}[\/latex],<\/div>\n<\/div>\n<\/div>\n<p>yielding<\/p>\n<div id=\"m42099-import-auto-id2171362\" class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.48.\"><span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x\u00a0<\/em><\/span>= 402 m.<\/span><\/div>\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\n<p>If we convert 402 m to miles, we find that the distance covered is very close to one quarter of a mile, the standard distance for drag racing. So the answer is reasonable. This is an impressive displacement in only 5.56 s, but top-notch dragsters can do a quarter mile in even less time than this.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>What else can we learn by examining the equation [latex]x={x}_{0}+{v}_{0}t+\\frac{1}{2}{\\text{at}}^{2}[\/latex]? We see that:<\/p>\n<div class=\"itemizedlist\">\n<ul class=\"itemizedlist\">\n<li class=\"listitem\">displacement depends on the square of the elapsed time when acceleration is not zero. In Example 3, the dragster covers only one fourth of the total distance in the first half of the elapsed time<\/li>\n<li class=\"listitem\">if acceleration is zero, then the initial velocity equals average velocity ([latex]{v}_{0}=\\bar{v}[\/latex]) and [latex]x={x}_{0}+{v}_{0}t+\\frac{1}{2}{\\text{at}}^{2}[\/latex] becomes <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x\u00a0<\/em><\/span>=\u00a0<span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>0\u00a0<\/sub>+\u00a0<span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0<\/sub><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span><\/li>\n<\/ul>\n<\/div>\n<div id=\"m42099-fs-id1164906443776\" class=\"example\" title=\"Example 2.11. Calculating Final Velocity: Dragsters\">\n<div class=\"textbox examples\">\n<h3>Example 4: Calculating Final Velocity: Dragsters<\/h3>\n<p>Calculate the final velocity of the dragster in Example 3 without using information about time.<\/p>\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\n<p>Draw a sketch.<\/p>\n<div id=\"m42099-import-auto-id4179116\" class=\"figure\" title=\"Figure 2.33.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 360px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101641\/Figure_02_04_02b.jpg\" alt=\"Acceleration vector arrow pointing toward the right, labeled twenty-six point zero meters per second squared. Initial velocity equals 0. Final velocity equals question mark.\" width=\"350\" height=\"175\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 8.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"title\"><\/div>\n<\/div>\n<p>The equation [latex]{v}^{2}={v}_{0}^{2}+2a\\left(x-{x}_{0}\\right)[\/latex] is ideally suited to this task because it relates velocities, acceleration, and displacement, and no time information is required.<\/p>\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\n<p>1. Identify the known values. We know that <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0\u00a0<\/sub>= 0<\/span>, since the dragster starts from rest. Then we note that <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span>\u2212<span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>0\u00a0<\/sub>= 402 m<\/span> (this was the answer in Example 3). Finally, the average acceleration was given to be <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a\u00a0<\/em><\/span>= 26.0 m\/s<sup>2<\/sup><\/span>.<\/p>\n<p>2. Plug the knowns into the equation[latex]{v}^{2}={v}_{0}^{2}+2a\\left(x-{x}_{0}\\right)[\/latex] and solve for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span>.<\/span><\/p>\n<div id=\"m42099-import-auto-id1680164\" class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.51.\"><span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sup>2\u00a0<\/sup>= 0 + 2(26.0 m\/s<sup>2<\/sup>)(402 m).<\/span><\/div>\n<p>Thus<\/p>\n<div id=\"m42099-import-auto-id2177794\" class=\"equation\" title=\"Equation 2.52.\">\n<div class=\"mediaobject\">\n<div id=\"import-auto-id2177794\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]{v}^{2}=2\\text{.}\\text{09}\\times {\\text{10}}^{4}{\\text{m}}^{2}{\\text{\/s}}^{2}[\/latex].<\/div>\n<div class=\"equation\" style=\"text-align: center;\" data-type=\"equation\"><\/div>\n<\/div>\n<\/div>\n<p>To get <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span>, we take the square root:<\/p>\n<div id=\"m42099-import-auto-id2177812\" class=\"equation\" title=\"Equation 2.53.\">\n<div class=\"mediaobject\">\n<div id=\"import-auto-id2177812\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]v=\\sqrt{2\\text{.}\\text{09}\\times {\\text{10}}^{4}{\\text{m}}^{2}{\\text{\/s}}^{2}}=\\text{145 m\/s}[\/latex].<\/div>\n<\/div>\n<\/div>\n<p><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/p>\n<p>145 m\/s is about 522 km\/h or about 324 mi\/h, but even this breakneck speed is short of the record for the quarter mile. Also, note that a square root has two values; we took the positive value to indicate a velocity in the same direction as the acceleration.<\/p>\n<\/div>\n<\/div>\n<p>An examination of the equation [latex]{v}^{2}={v}_{0}^{2}+2a\\left(x-{x}_{0}\\right)[\/latex] can produce further insights into the general relationships among physical quantities:<\/p>\n<div class=\"itemizedlist\">\n<ul class=\"itemizedlist\">\n<li class=\"listitem\">The final velocity depends on how large the acceleration is and the distance over which it acts<\/li>\n<li class=\"listitem\">For a fixed deceleration, a car that is going twice as fast doesn\u2019t simply stop in twice the distance\u2014it takes much further to stop. (This is why we have reduced speed zones near schools.)<\/li>\n<\/ul>\n<\/div>\n<h2 id=\"m42099-fs-id1164906446591\"><span class=\"cnx-gentext-section cnx-gentext-t\">Putting Equations Together<\/span><\/h2>\n<p>In the following examples, we further explore one-dimensional motion, but in situations requiring slightly more algebraic manipulation. The examples also give insight into problem-solving techniques. The box below provides easy reference to the equations needed.<\/p>\n<div class=\"textbox shaded\">\n<h3 class=\"title\"><strong><span class=\"cnx-gentext-tip-t\">Summary of Kinematic Equations (constant <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span>)<\/span><\/strong><\/h3>\n<div class=\"body\">\n<div id=\"m42099-import-auto-id1771742\" class=\"equation\" title=\"Equation 2.54.\">\n<div class=\"mediaobject\">\n<div id=\"import-auto-id1771742\" class=\"equation\" data-type=\"equation\">\n<div id=\"import-auto-id1771742\" class=\"equation\" data-type=\"equation\">[latex]x={x}_{0}+\\bar{v}t[\/latex]<\/div>\n<div id=\"import-auto-id4178996\" class=\"equation\" data-type=\"equation\">[latex]\\bar{v}=\\frac{{v}_{0}+v}{2}[\/latex]<\/div>\n<div id=\"import-auto-id1680040\" class=\"equation\" data-type=\"equation\">[latex]v={v}_{0}+\\text{at}[\/latex]<\/div>\n<div id=\"import-auto-id1680037\" class=\"equation\" data-type=\"equation\">[latex]x={x}_{0}+{v}_{0}t+\\frac{1}{2}{\\text{at}}^{2}[\/latex]<\/div>\n<div id=\"import-auto-id2178979\" class=\"equation\" data-type=\"equation\">[latex]{v}^{2}={v}_{0}^{2}+2a\\left(x-{x}_{0}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"m42099-fs-id1164906424692\" class=\"example\" title=\"Example 2.12. Calculating Displacement: How Far Does a Car Go When Coming to a Halt?\">\n<div class=\"textbox examples\">\n<h3>Example 5. Calculating Displacement: How Far Does a Car Go When Coming to a Halt?<\/h3>\n<p>On dry concrete, a car can decelerate at a rate of <span class=\"token\">7.00 m\/s<sup>2<\/sup><\/span>, whereas on wet concrete it can decelerate at only <span class=\"token\">5.00 m\/s<sup>2<\/sup><\/span>. Find the distances necessary to stop a car moving at 30.0 m\/s (about 110 km\/h) (a) on dry concrete and (b) on wet concrete. (c) Repeat both calculations, finding the displacement from the point where the driver sees a traffic light turn red, taking into account his reaction time of 0.500 s to get his foot on the brake.<\/p>\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\n<p>Draw a sketch.<\/p>\n<div id=\"m42099-import-auto-id2174020\" class=\"figure\" title=\"Figure 2.34.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 360px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101649\/Figure_02_04_02c.jpg\" alt=\"Initial velocity equals thirty meters per second. Final velocity equals 0. Acceleration dry equals negative 7 point zero zero meters per second squared. Acceleration wet equals negative 5 point zero zero meters per second squared.\" width=\"350\" height=\"253\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 9.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"title\"><\/div>\n<\/div>\n<p>In order to determine which equations are best to use, we need to list all of the known values and identify exactly what we need to solve for. We shall do this explicitly in the next several examples, using tables to set them off.<\/p>\n<h4><span class=\"bold\"><strong>Solution for (a)<\/strong><\/span><\/h4>\n<p>1. Identify the knowns and what we want to solve for. We know that <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0\u00a0<\/sub>= 30.0 m\/s<\/span>; <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v\u00a0<\/em><\/span>= 0<\/span>; a = -7.00 m\/s<sup>2<\/sup>\u00a0(<span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span> is negative because it is in a direction opposite to velocity). We take <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>0<\/sub><\/span> to be 0. We are looking for displacement <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>, or <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span>\u2212<span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>0<\/sub><\/span>.<\/p>\n<p>2. Identify the equation that will help up solve the problem. The best equation to use is<\/p>\n<div id=\"m42099-import-auto-id2180580\" class=\"equation\" title=\"Equation 2.59.\">\n<div class=\"mediaobject\">\n<div id=\"import-auto-id2180580\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]{v}^{2}={v}_{0}^{2}+2a\\left(x-{x}_{0}\\right)[\/latex].<\/div>\n<div class=\"equation\" style=\"text-align: center;\" data-type=\"equation\"><\/div>\n<\/div>\n<\/div>\n<p>This equation is best because it includes only one unknown, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>. We know the values of all the other variables in this equation. (There are other equations that would allow us to solve for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>, but they require us to know the stopping time, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span>, which we do not know. We could use them but it would entail additional calculations.)<\/p>\n<p>3. Rearrange the equation to solve for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>.<\/p>\n<div id=\"m42099-import-auto-id4179294\" class=\"equation\" title=\"Equation 2.60.\">\n<div class=\"mediaobject\">\n<div id=\"import-auto-id4179294\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]x-{x}_{0}=\\frac{{v}^{2}-{v}_{0}^{2}}{2a}[\/latex]<\/div>\n<div class=\"equation\" style=\"text-align: center;\" data-type=\"equation\"><\/div>\n<\/div>\n<\/div>\n<p>4. Enter known values.<\/p>\n<div id=\"m42099-import-auto-id2293184\" class=\"equation\" title=\"Equation 2.61.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]x - 0=\\frac{{0}^{2}-{\\left(\\text{30}\\text{.}\\text{0 m\/s}\\right)}^{2}}{2\\left(-7\\text{.}{\\text{00 m\/s}}^{2}\\right)}[\/latex]<\/div>\n<div class=\"mediaobject\"><\/div>\n<\/div>\n<p>Thus,<\/p>\n<div id=\"m42099-import-auto-id2293155\" class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.62.\"><span class=\"token\"> <span class=\"emphasis mathml-mi\"><em>x<\/em><\/span> = 64.3 m on dry concrete.<\/span><\/div>\n<h4><span class=\"bold\"><strong>Solution for (b)<\/strong><\/span><\/h4>\n<p>This part can be solved in exactly the same manner as Part A. The only difference is that the deceleration is <span class=\"token\">\u20135.00 m\/s<sup>2<\/sup><\/span>. The result is<\/p>\n<div id=\"m42099-import-auto-id2178635\" class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.63.\"><span class=\"token\"> <span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub> wet <\/sub> = 90.0 m on wet concrete.<\/span><\/div>\n<h4><span class=\"bold\"><strong>Solution for (c)<\/strong><\/span><\/h4>\n<p>Once the driver reacts, the stopping distance is the same as it is in Parts A and B for dry and wet concrete. So to answer this question, we need to calculate how far the car travels during the reaction time, and then add that to the stopping time. It is reasonable to assume that the velocity remains constant during the driver\u2019s reaction time.<\/p>\n<p>1. Identify the knowns and what we want to solve for. We know that [latex]\\bar{v}=30.0 \\text{ m\/s}[\/latex]; <em>t<\/em><sub>reaction<\/sub> = 0.500 s; <em>a<\/em><sub>reaction<\/sub> = 0. We take<i>\u00a0x<sub>0-<\/sub><\/i><sub>reaction\u00a0<\/sub><em>=\u00a0<\/em>to be 0. We are looking for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>reaction<\/sub><\/span>.<\/p>\n<p>2. Identify the best equation to use.\u00a0[latex]x={x}_{0}+\\bar{v}t[\/latex]\u00a0works well because the only unknown value is <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>, which is what we want to solve for.<\/p>\n<p>3. Plug in the knowns to solve the equation.<\/p>\n<div id=\"m42099-import-auto-id2175306\" class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.64.\"><span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x\u00a0<\/em><\/span>= 0+(30.0 m\/s)(0.500 s)=15.0 m.<\/span><\/div>\n<div class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.64.\"><\/div>\n<p>This means the car travels 15.0 m while the driver reacts, making the total displacements in the two cases of dry and wet concrete 15.0 m greater than if he reacted instantly.<\/p>\n<p>4. Add the displacement during the reaction time to the displacement when braking.<\/p>\n<div id=\"m42099-import-auto-id1658817\" class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.65.\"><span class=\"token\"> <span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub> braking <\/sub> + <span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub> reaction <\/sub> = <span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub> total<\/sub><\/span><\/div>\n<div class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.65.\"><\/div>\n<div class=\"orderedlist\">\n<ol class=\"orderedlist\" type=\"a\">\n<li class=\"listitem\">64.3 m + 15.0 m = 79.3 m when dry<\/li>\n<li class=\"listitem\">90.0 m + 15.0 m = 105 m when wet<\/li>\n<\/ol>\n<\/div>\n<div id=\"m42099-import-auto-id1658840\" class=\"figure\" title=\"Figure 2.35.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 610px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101654\/Figure_02_04_03.jpg\" alt=\"Diagram showing the various braking distances necessary for stopping a car. With no reaction time considered, braking distance is 64 point 3 meters on a dry surface and 90 meters on a wet surface. With reaction time of 0 point 500 seconds, braking distance is 79 point 3 meters on a dry surface and 105 meters on a wet surface.\" width=\"600\" height=\"379\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 10. The distance necessary to stop a car varies greatly, depending on road conditions and driver reaction time. Shown here are the braking distances for dry and wet pavement, as calculated in this example, for a car initially traveling at 30.0 m\/s. Also shown are the total distances traveled from the point where the driver first sees a light turn red, assuming a 0.500 s reaction time.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\n<p>The displacements found in this example seem reasonable for stopping a fast-moving car. It should take longer to stop a car on wet rather than dry pavement. It is interesting that reaction time adds significantly to the displacements. But more important is the general approach to solving problems. We identify the knowns and the quantities to be determined and then find an appropriate equation. There is often more than one way to solve a problem. The various parts of this example can in fact be solved by other methods, but the solutions presented above are the shortest.<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 6. Calculating Time: A Car Merges into Traffic<\/h3>\n<div class=\"body\">\n<p>Suppose a car merges into freeway traffic on a 200-m-long ramp. If its initial velocity is 10.0 m\/s and it accelerates at <span class=\"token\">2.00 m\/s<sup>2<\/sup><\/span>, how long does it take to travel the 200 m up the ramp? (Such information might be useful to a traffic engineer.)<\/p>\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\n<p>Draw a sketch.<\/p>\n<div id=\"m42099-import-auto-id2296882\" class=\"figure\" title=\"Figure 2.36.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 260px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101656\/Figure_02_04_03a.jpg\" alt=\"A line segment with ends labeled x subs zero equals zero and x = two hundred. Above the line segment, the equation t equals question mark indicates that time is unknown. Three vectors, all pointing in the direction of x equals 200, represent the other knowns and unknowns. They are labeled v sub zero equals ten point zero meters per second, v equals question mark, and a equals two point zero zero meters per second squared.\" width=\"250\" height=\"289\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 11.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"title\"><\/div>\n<\/div>\n<p>We are asked to solve for the time <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span>. As before, we identify the known quantities in order to choose a convenient physical relationship (that is, an equation with one unknown, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span>).<\/p>\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\n<p>1. Identify the knowns and what we want to solve for. We know that <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0\u00a0<\/sub>= 10 m\/s<\/span>; <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a\u00a0<\/em><\/span>= 2.00 m\/s<sup>2<\/sup><\/span>; and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x\u00a0<\/em><\/span>= 200 m<\/span>.<\/p>\n<p>2. We need to solve for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span>. Choose the best equation. [latex]x={x}_{0}+{v}_{0}t+\\frac{1}{2}{{at}}^{2}[\/latex]\u00a0works best because the only unknown in the equation is the variable <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> for which we need to solve.<\/p>\n<p>3. We will need to rearrange the equation to solve for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span>. In this case, it will be easier to plug in the knowns first.<\/p>\n<p style=\"text-align: center;\">[latex]\\text{200 m}=\\text{0 m}+\\left(\\text{10}\\text{.}\\text{0 m\/s}\\right)t+\\frac{1}{2}\\left(2\\text{.}{\\text{00 m\/s}}^{2}\\right){t}^{2}[\/latex]<\/p>\n<p>4. Simplify the equation. The units of meters (m) cancel because they are in each term. We can get the units of seconds (s) to cancel by taking <em>t<\/em> = <em>ts<\/em>, where <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> is the magnitude of time and s is the unit. Doing so leaves<\/p>\n<div id=\"m42099-eip-635\" class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.67.\"><span class=\"token\">200 = 10<span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>+\u00a0<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sup>2<\/sup>.<\/span><\/div>\n<div class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.67.\"><\/div>\n<p>5. Use the quadratic formula to solve for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span><span class=\"emphasis\"><em>. <\/em><\/span><\/p>\n<p>(a) Rearrange the equation to get 0 on one side of the equation.<\/p>\n<div id=\"m42099-import-auto-id1680208\" class=\"equation\" style=\"text-align: center;\" title=\"Equation 2.68.\"><span class=\"token\"> <span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sup> 2 <\/sup> + 10<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span> \u2212 200 = 0 <\/span><\/div>\n<p>This is a quadratic equation of the form<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{at}}^{2}+\\text{bt}+c=0[\/latex]<\/p>\n<p>where the constants are <em>a<\/em> = 1.00, <em>b<\/em> = 10.0 and <em>c<\/em> = -200.<\/p>\n<p>(b) Its solutions are given by the quadratic formula:<\/p>\n<div id=\"import-auto-id2367246\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]t=\\frac{-b\\pm \\sqrt{{b}^{2}-4\\text{ac}}}{2a}[\/latex]<\/div>\n<div class=\"equation\" data-type=\"equation\"><\/div>\n<p>This yields two solutions for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span>, which are<\/p>\n<p style=\"text-align: center;\"><em>t<\/em> = 10.0 and -20.0.<\/p>\n<p>In this case, then, the time is <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>=\u00a0<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> in seconds, or<\/p>\n<p style=\"text-align: center;\"><em>t<\/em> = 10.0 s and -20.0 s.<\/p>\n<p>A negative value for time is unreasonable, since it would mean that the event happened 20 s before the motion began. We can discard that solution. Thus,<\/p>\n<p style=\"text-align: center;\"><em>t<\/em> = 10.0 s.<\/p>\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\n<p>Whenever an equation contains an unknown squared, there will be two solutions. In some problems both solutions are meaningful, but in others, such as the above, only one solution is reasonable. The 10.0 s answer seems reasonable for a typical freeway on-ramp.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>With the basics of kinematics established, we can go on to many other interesting examples and applications. In the process of developing kinematics, we have also glimpsed a general approach to problem solving that produces both correct answers and insights into physical relationships. <a class=\"link\" title=\"2.6. Problem-Solving Basics for One-Dimensional Kinematics\" href=\".\/chapter\/2-6-problem-solving-basics-for-one-dimensional-kinematics\/\" target=\"_blank\">Problem-Solving Basics<\/a> discusses problem-solving basics and outlines an approach that will help you succeed in this invaluable task.<\/p>\n<div id=\"m42099-fs-id1164906508057\" class=\"note\">\n<div class=\"textbox learning-objectives\">\n<h3><strong><span class=\"cnx-gentext-tip-t\">Making Connections: Take-Home Experiment\u2014Breaking News<\/span><\/strong><\/h3>\n<div class=\"body\">We have been using SI units of meters per second squared to describe some examples of acceleration or deceleration of cars, runners, and trains. To achieve a better feel for these numbers, one can measure the braking deceleration of a car doing a slow (and safe) stop. Recall that, for average acceleration, [latex]\\bar{a}=\\Delta v\/ \\Delta t[\/latex]. While traveling in a car, slowly apply the brakes as you come up to a stop sign. Have a passenger note the initial speed in miles per hour and the time taken (in seconds) to stop. From this, calculate the deceleration in miles per hour per second. Convert this to meters per second squared and compare with other decelerations mentioned in this chapter. Calculate the distance traveled in braking.<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Check Your Understanding<\/h3>\n<div id=\"m42099-fs-id1164906508057\" class=\"note\">\n<p>A manned rocket accelerates at a rate of <span class=\"token\">20 m\/s<sup>2<\/sup><\/span> during launch. How long does it take the rocket reach a velocity of 400 m\/s?<\/p>\n<\/div>\n<div id=\"m42099-fs-id1164906458555\" class=\"solution\">\n<h4 class=\"title\"><strong><span class=\"epub-only pre-text\">Solution<\/span><\/strong><\/h4>\n<div class=\"body\">\n<p>To answer this, choose an equation that allows you to solve for time <span class=\"emphasis\"><em><span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span><\/em><\/span>, given only <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0<\/sub><\/span>, and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span>.<\/p>\n<p style=\"text-align: center;\">[latex]v={v}_{0}+{at}[\/latex]<\/p>\n<p>Rearrange to solve for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span><span class=\"emphasis\"><em>. <\/em><\/span><\/p>\n<div id=\"m42099-import-auto-id2168930\" class=\"equation\" title=\"Equation 2.75.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]t=\\frac{v-v{}_{0}\\text{}}{a}=\\frac{\\text{400 m\/s}-\\text{0 m\/s}}{{\\text{20 m\/s}}^{2}}=\\text{20 s}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>Section Summary<\/h2>\n<ul>\n<li>To simplify calculations we take acceleration to be constant, so that [latex]\\bar{a}=a[\/latex] at all times.<\/li>\n<li>We also take initial time to be zero.<\/li>\n<li><span style=\"color: #000000;\">Initial position and velocity are given a subscript 0; final values have no subscript. Thus,<\/span><br \/>\n[latex]\\begin{cases}{\\Delta}{t} &=& t \\\\{\\Delta}{x} &=& x-{{x}_{0}}\\\\{\\Delta}{v} &=& v-{{v}_{0}}\\end{cases}[\/latex]<\/li>\n<li>The following kinematic equations for motion with constant [latex]a[\/latex] are useful:\n<div id=\"import-auto-id1771742\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]x={x}_{0}+\\bar{v}t[\/latex]<\/div>\n<div id=\"import-auto-id4178996\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\bar{v}=\\frac{{v}_{0}+v}{2}[\/latex]<\/div>\n<div id=\"import-auto-id1680040\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]v={v}_{0}+\\text{at}[\/latex]<\/div>\n<div id=\"import-auto-id1680037\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]x={x}_{0}+{v}_{0}t+\\frac{1}{2}{\\text{at}}^{2}[\/latex]<\/div>\n<div id=\"import-auto-id2178979\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]{v}^{2}={v}_{0}^{2}+2a\\left(x-{x}_{0}\\right)[\/latex]<\/div>\n<p>In vertical motion, <em>y<\/em>\u00a0is substituted for <em>x<\/em>.<\/li>\n<\/ul>\n<div class=\"textbox exercises\">\n<h3>Problems &amp; Exercises<\/h3>\n<p>1. An Olympic-class sprinter starts a race with an acceleration of 4.50 m\/s<sup>2<\/sup>. (a) What is her speed 2.40 s later? (b) Sketch a graph of her position vs. time for this period.<\/p>\n<p>2. A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10 \u00d7 10<sup>4<\/sup> m\/s<sup>2<\/sup>, and 1.85 ms (1 ms = 10-3 s) elapses from the time the ball first touches the mitt until it stops, what was the initial velocity of the ball?<\/p>\n<p>3. A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 6.20 \u00d7 10<sup>5\u00a0<\/sup>m\/s<sup>2<\/sup> for\u00a08.10 \u00d7 10<sup>-4 <\/sup>s.\u00a0What is its muzzle velocity (that is, its final velocity)?<\/p>\n<p>4.\u00a0(a) A light-rail commuter train accelerates at a rate of <span class=\"token\">1.35 m\/s<sup>2<\/sup><\/span>. How long does it take to reach its top speed of 80.0 km\/h, starting from rest?\u00a0(b) The same train ordinarily decelerates at a rate of <span class=\"token\">1.65 m\/s<sup>2<\/sup><\/span>. How long does it take to come to a stop from its top speed?\u00a0(c) In emergencies the train can decelerate more rapidly, coming to rest from 80.0 km\/h in 8.30 s. What is its emergency deceleration in <span class=\"token\">m\/s<sup>2<\/sup><\/span>?<\/p>\n<p>5.\u00a0While entering a freeway, a car accelerates from rest at a rate of <span class=\"token\">2.40 m\/s<sup>2<\/sup><\/span> for 12.0 s.\u00a0(a) Draw a sketch of the situation.\u00a0(b) List the knowns in this problem.\u00a0(c) How far does the car travel in those 12.0 s?\u00a0To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, check your units, and discuss whether the answer is reasonable.\u00a0(d) What is the car\u2019s final velocity? Solve for this unknown in the same manner as in part (c), showing all steps explicitly.<\/p>\n<p>6. At the end of a race, a runner decelerates from a velocity of 9.00 m\/s at a rate of <span class=\"token\">2.00 m\/s<sup>2<\/sup><\/span>.\u00a0(a) How far does she travel in the next 5.00 s?\u00a0(b) What is her final velocity?\u00a0(c) Evaluate the result. Does it make sense?<\/p>\n<p>7. <strong>Professional Application:\u00a0<\/strong>Blood is accelerated from rest to 30.0 cm\/s in a distance of 1.80 cm by the left ventricle of the heart. (a) Make a sketch of the situation. (b) List the knowns in this problem. (c) How long does the acceleration take? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units. (d) Is the answer reasonable when compared with the time for a heartbeat?<\/p>\n<p>8. In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m\/s to 40.0 m\/s in the same direction. If this shot takes 3.33 \u00d7 10<sup>-2<\/sup>, calculate the distance over which the puck accelerates.<\/p>\n<p>9. A powerful motorcycle can accelerate from rest to 26.8 m\/s (100 km\/h) in only 3.90 s. (a) What is its average acceleration? (b) How far does it travel in that time?<\/p>\n<p>10. Freight trains can produce only relatively small accelerations and decelerations. (a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m\/s<sup>2<\/sup> for 8.00 min, starting with an initial velocity of 4.00 m\/s? (b) If the train can slow down at a rate of 0.550 m\/s<sup>2<\/sup>, how long will it take to come to a stop from this velocity? (c) How far will it travel in each case?<\/p>\n<p>11.\u00a0A fireworks shell is accelerated from rest to a velocity of 65.0 m\/s over a distance of 0.250 m. (a) How long did the acceleration last? (b) Calculate the acceleration.<\/p>\n<p>12. A swan on a lake gets airborne by flapping its wings and running on top of the water. (a) If the swan must reach a velocity of 6.00 m\/s to take off and it accelerates from rest at an average rate of 0.350 m\/s<sup>2<\/sup>, how far will it travel before becoming airborne? (b) How long does this take?<\/p>\n<p>13. <strong>Professional Application:\u00a0<\/strong>A woodpecker\u2019s brain is specially protected from large decelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker\u2019s head comes to a stop from an initial velocity of 0.600 m\/s in a distance of only 2.00 mm. (a) Find the acceleration in m\/s<sup>2<\/sup> and in multiples of <em>g<\/em> (<em>g<\/em> = 9.80 m\/s<sup>2<\/sup>. (b) Calculate the stopping time. (c) The tendons cradling the brain stretch, making its stopping distance 4.50 mm (greater than the head and, hence, less deceleration of the brain). What is the brain\u2019s deceleration, expressed in multiples of\u00a0<em>g<\/em>?<\/p>\n<p>14. An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m\/s and comes to a full stop after compressing the padding and his body 0.350 m. (a) What is his deceleration? (b) How long does the collision last?<\/p>\n<p>15. In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no parachute to escape certain death. Some fell about 20,000 feet (6000 m), and some of them survived, with few life-threatening injuries. For these lucky pilots, the tree branches and snow drifts on the ground allowed their deceleration to be relatively small. If we assume that a pilot\u2019s speed upon impact was 123 mph (54 m\/s), then what was his deceleration? Assume that the trees and snow stopped him over a distance of 3.0 m.<\/p>\n<p>16. Consider a grey squirrel falling out of a tree to the ground. (a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel\u2019s velocity just before hitting the ground, assuming it fell from a height of 3.0 m. (b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem.<\/p>\n<p>17. An express train passes through a station. It enters with an initial velocity of 22.0 m\/s and decelerates at a rate of [latex]0text{.}{text{150 m\/s}}^{2}[\/latex] as it goes through. The station is 210 m long. (a) How long is the nose of the train in the station? (b) How fast is it going when the nose leaves the station? (c) If the train is 130 m long, when does the end of the train leave the station? (d) What is the velocity of the end of the train as it leaves?<\/p>\n<p>18. Dragsters can actually reach a top speed of 145 m\/s in only 4.45 s\u2014considerably less time than given in Example 2.10 and Example 2.11. (a) Calculate the average acceleration for such a dragster. (b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any information on time. (c) Why is the final velocity greater than that used to find the average acceleration? Hint: Consider whether the assumption of constant acceleration is valid for a dragster. If not, discuss whether the acceleration would be greater at the beginning or end of the run and what effect that would have on the final velocity.<\/p>\n<p>19. A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m\/s and accelerates at the rate of 0.500 m\/s<sup>2<\/sup>\u00a0for 7.00 s. (a) What is his final velocity? (b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save? (c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m\/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?<\/p>\n<p>20. In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, with a maximum speed of 183.58 mi\/h. The one-way course was 5.00 mi long. Acceleration rates are often described by the time it takes to reach 60.0 mi\/h from rest. If this time was 4.00 s, and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course?<\/p>\n<p>21. (a) A world record was set for the men\u2019s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt &#8220;coasted&#8221; across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration. (b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Selected Solutions to Problems &amp; Exercises<\/h3>\n<p>1.\u00a010.8 m\/s<\/p>\n<p>(b)<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1322\/2015\/12\/03205527\/unnumbered_art_p44.jpg\" alt=\"Line graph of position in meters versus time in seconds. The line begins at the origin and is concave up, with its slope increasing over time.\" width=\"350\" height=\"441\" data-media-type=\"image\/jpg\" \/><\/p>\n<p>2.\u00a038.9 m\/s (about 87 miles per hour)<\/p>\n<p>4. (a) 16.5 s (b) 13.5 s (c) -2.68 m\/s<sup>2<\/sup><\/p>\n<p>6.\u00a0(a)\u00a020.0 m (b) -1.00 m\/s\u00a0(c) This result does not really make sense. If the runner starts at 9.00 m\/s and decelerates at 2.00 m\/s<sup>2<\/sup>, then she will have stopped after 4.50 s. If she continues to decelerate, she will be running backwards.<\/p>\n<p>8. 0.799 m<\/p>\n<p>10. (a) 28.0 m\/s (b) 50.9 s\u00a0(c) 7.68 km to accelerate and 713 m to decelerate<\/p>\n<p>12. (a) 51.4 m (b) 17.1 s<\/p>\n<p>14. (a) -80\u00a0m\/s<sup>2\u00a0<\/sup>(b)\u00a09.33 \u00d7 10<sup>&#8211;<\/sup><sup>2<\/sup> s<\/p>\n<p>16. (a) 7.7 m\/s (b) -15 \u00d7 10<sup>2<\/sup>\u00a0m\/s<sup>2\u00a0<\/sup>his is about 3 times the deceleration of the pilots, who were falling from thousands of meters high!<\/p>\n<p>18.\u00a0(a)\u00a036.2\u00a0m\/s<sup>2\u00a0<\/sup>(b)\u00a0162 m\/s\u00a0(c) v &gt; v<sub>max<\/sub>, because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears, and would have a greater acceleration in first gear than second gear than third gear, etc. The acceleration would be greatest at the beginning, so it would not be accelerating at 32 m\/s<sup>2<\/sup> during the last few meters, but substantially less, and the final velocity would be less than 162 m\/s.<\/p>\n<p>20.\u00a0104 s<\/p>\n<p>21.\u00a0(a)\u00a0<em>v<\/em> = 12\/2 m\/s; <em>a<\/em> = 4.07 m\/s<sup>2\u00a0<\/sup>(b)<em> v<\/em> = 11.2 m\/s<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-285\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Physics. <strong>Authored by<\/strong>: OpenStax College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/031da8d3-b525-429c-80cf-6c8ed997733a\/College_Physics\">http:\/\/cnx.org\/contents\/031da8d3-b525-429c-80cf-6c8ed997733a\/College_Physics<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Located at License<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":1,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"College Physics\",\"author\":\"OpenStax College\",\"organization\":\"\",\"url\":\"http:\/\/cnx.org\/contents\/031da8d3-b525-429c-80cf-6c8ed997733a\/College_Physics\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Located at License\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-285","chapter","type-chapter","status-publish","hentry"],"part":7456,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/chapters\/285","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/wp\/v2\/users\/1"}],"version-history":[{"count":68,"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/chapters\/285\/revisions"}],"predecessor-version":[{"id":12295,"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/chapters\/285\/revisions\/12295"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/parts\/7456"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/chapters\/285\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/wp\/v2\/media?parent=285"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/chapter-type?post=285"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/wp\/v2\/contributor?post=285"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/wp\/v2\/license?post=285"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}