{"id":353,"date":"2018-02-06T15:33:39","date_gmt":"2018-02-06T15:33:39","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/?post_type=chapter&#038;p=353"},"modified":"2018-07-04T14:56:18","modified_gmt":"2018-07-04T14:56:18","slug":"3-2-instantaneous-velocity-and-speed","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/chapter\/3-2-instantaneous-velocity-and-speed\/","title":{"raw":"3.2 Instantaneous Velocity and Speed","rendered":"3.2 Instantaneous Velocity and Speed"},"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<ul>\r\n \t<li>Explain the difference between average velocity and instantaneous velocity.<\/li>\r\n \t<li>Describe the difference between velocity and speed.<\/li>\r\n \t<li>Calculate the instantaneous velocity given the mathematical equation for the velocity.<\/li>\r\n \t<li>Calculate the speed given the instantaneous velocity.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1168326781535\">We have now seen how to calculate the average velocity between two positions. However, since objects in the real world move continuously through space and time, we would like to find the velocity of an object at any single point. We can find the velocity of the object anywhere along its path by using some fundamental principles of calculus. This section gives us better insight into the physics of motion and will be useful in later chapters.<\/p>\r\n\r\n<div id=\"fs-id1168329168511\" class=\"bc-section section\">\r\n<h3>Instantaneous Velocity<\/h3>\r\n<p id=\"fs-id1168329339669\">The quantity that tells us how fast an object is moving anywhere along its path is the<strong> instantaneous velocity<\/strong>, usually called simply <em>velocity<\/em>. It is the average velocity between two points on the path in the limit that the time (and therefore the displacement) between the two points approaches zero. To illustrate this idea mathematically, we need to express position <em>x<\/em> as a continuous function of <em>t<\/em> denoted by <em>x<\/em>(<em>t<\/em>). The expression for the average velocity between two points using this notation is $$ \\overset{\\text{\u2013}}{v}=\\frac{x({t}_{2})-x({t}_{1})}{{t}_{2}-{t}_{1}}$$. To find the instantaneous velocity at any position, we let $$ {t}_{1}=t $$ and $$ {t}_{2}=t+\\text{\u0394}t$$. After inserting these expressions into the equation for the average velocity and taking the limit as $$ \\text{\u0394}t\\to 0$$, we find the expression for the instantaneous velocity:<\/p>\r\n\r\n<div id=\"fs-id1168329127174\" class=\"unnumbered\">$$v(t)=\\underset{\\text{\u0394}t\\to 0}{\\text{lim}}\\frac{x(t+\\text{\u0394}t)-x(t)}{\\text{\u0394}t}=\\frac{dx(t)}{dt}.$$<\/div>\r\n<div id=\"fs-id1168329192792\">\r\n<div>Instantaneous Velocity<\/div>\r\n<p id=\"fs-id1168326802131\">The instantaneous velocity of an object is the limit of the average velocity as the elapsed time approaches zero, or the derivative of <em>x<\/em> with respect to <em>t<\/em>:<\/p>\r\n\r\n<div id=\"fs-id1168329153584\">$$v(t)=\\frac{d}{dt}x(t).$$<\/div>\r\n<\/div>\r\n<p id=\"fs-id1168326795408\">Like average velocity, instantaneous velocity is a vector with dimension of length per time. The instantaneous velocity at a specific time point $$ {t}_{0} $$ is the rate of change of the position function, which is the slope of the position function $$ x(t) $$ at $$ {t}_{0}$$. <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_03_02_InstVel\">(Figure)<\/a> shows how the average velocity $$ \\overset{\\text{\u2013}}{v}=\\frac{\\text{\u0394}x}{\\text{\u0394}t} $$ between two times approaches the instantaneous velocity at $$ {t}_{0}. $$ The instantaneous velocity is shown at time $$ {t}_{0}$$, which happens to be at the maximum of the position function. The slope of the position graph is zero at this point, and thus the instantaneous velocity is zero. At other times, $$ {t}_{1},{t}_{2}$$, and so on, the instantaneous velocity is not zero because the slope of the position graph would be positive or negative. If the position function had a minimum, the slope of the position graph would also be zero, giving an instantaneous velocity of zero there as well. Thus, the zeros of the velocity function give the minimum and maximum of the position function.<\/p>\r\n\r\n<div id=\"CNX_UPhysics_03_02_InstVel\" class=\"wp-caption aligncenter\">[caption id=\"\" align=\"aligncenter\" width=\"414\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184932\/CNX_UPhysics_03_02_InstVel.jpg\" alt=\"Graph shows position plotted versus time. Position increases from t1 to t2 and reaches maximum at t0. It decreases to at and continues to decrease at t4. The slope of the tangent line at t0 is indicated as the instantaneous velocity.\" width=\"414\" height=\"352\" \/> <strong>Figure 3.6<\/strong> In a graph of position versus time, the instantaneous velocity is the slope of the tangent line at a given point. The average velocities $$ \\overset{\\text{\u2013}}{v}=\\frac{\\text{\u0394}x}{\\text{\u0394}t}=\\frac{{x}_{\\text{f}}-{x}_{\\text{i}}}{{t}_{\\text{f}}-{t}_{\\text{i}}} $$ between times $$ \\text{\u0394}t={t}_{6}-{t}_{1},\\text{\u0394}t={t}_{5}-{t}_{2},\\text{and}\\,\\text{\u0394}t={t}_{4}-{t}_{3} $$ are shown. When $$ \\text{\u0394}t\\to 0$$, the average velocity approaches the instantaneous velocity at $$ t={t}_{0}$$.[\/caption]<\/div>\r\n<div id=\"fs-id1168329165028\" class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\n<h4>Finding Velocity from a Position-Versus-Time Graph<\/h4>\r\nGiven the position-versus-time graph of <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_03_02_PosGraph\">(Figure)<\/a>, find the velocity-versus-time graph.\r\n<div id=\"CNX_UPhysics_03_02_PosGraph\" class=\"wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"446\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184935\/CNX_UPhysics_03_02_PosGraph.jpg\" alt=\"Graph shows position in kilometers plotted as a function of time at minutes. It starts at the origin, reaches 0.5 kilometers at 0.5 minutes, remains constant between 0.5 and 0.9 minutes, and decreases to 0 at 2.0 minutes.\" width=\"446\" height=\"421\" \/> <strong>Figure 3.7<\/strong> The object starts out in the positive direction, stops for a short time, and then reverses direction, heading back toward the origin. Notice that the object comes to rest instantaneously, which would require an infinite force. Thus, the graph is an approximation of motion in the real world. (The concept of force is discussed in Newton\u2019s Laws of Motion.)[\/caption]\r\n\r\n<\/div>\r\n<h4>Strategy<\/h4>\r\nThe graph contains three straight lines during three time intervals. We find the velocity during each time interval by taking the slope of the line using the grid.\r\n<h4>Solution<\/h4>\r\n<p id=\"fs-id1168329296202\">[reveal-answer q=\"508307\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"508307\"]Time interval 0 s to 0.5 s: $$ \\overset{\\text{\u2013}}{v}=\\frac{\\text{\u0394}x}{\\text{\u0394}t}=\\frac{0.5\\,\\text{m}-0.0\\,\\text{m}}{0.5\\,\\text{s}-0.0\\,\\text{s}}=1.0\\,\\text{m\/s}$$<\/p>\r\nTime interval 0.5 s to 1.0 s: $$ \\overset{\\text{\u2013}}{v}=\\frac{\\text{\u0394}x}{\\text{\u0394}t}=\\frac{0.0\\,\\text{m}-0.0\\,\\text{m}}{1.0\\,\\text{s}-0.5\\,\\text{s}}=0.0\\,\\text{m\/s}$$\r\n\r\nTime interval 1.0 s to 2.0 s: $$ \\overset{\\text{\u2013}}{v}=\\frac{\\text{\u0394}x}{\\text{\u0394}t}=\\frac{0.0\\,\\text{m}-0.5\\,\\text{m}}{2.0\\,\\text{s}-1.0\\,\\text{s}}=-0.5\\,\\text{m\/s}$$\r\n\r\nThe graph of these values of velocity versus time is shown in (Figure).\r\n<div id=\"CNX_UPhysics_03_02_VelGraph\" class=\"wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"414\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184938\/CNX_UPhysics_03_02_VelGraph.jpg\" alt=\"Graph shows velocity in meters per second plotted as a function of time at seconds. The velocity is 1 meter per second between 0 and 0.5 seconds, zero between 0.5 and 1.0 seconds, and -0.5 between 1.0 and 2.0 seconds.\" width=\"414\" height=\"315\" \/> <strong>Figure 3.8<\/strong> The velocity is positive for the first part of the trip, zero when the object is stopped, and negative when the object reverses direction.[\/caption]\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n<h4>Significance<\/h4>\r\nDuring the time interval between 0 s and 0.5 s, the object\u2019s position is moving away from the origin and the position-versus-time curve has a positive slope. At any point along the curve during this time interval, we can find the instantaneous velocity by taking its slope, which is +1 m\/s, as shown in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_03_02_VelGraph\">(Figure)<\/a>. In the subsequent time interval, between 0.5 s and 1.0 s, the position doesn\u2019t change and we see the slope is zero. From 1.0 s to 2.0 s, the object is moving back toward the origin and the slope is \u22120.5 m\/s. The object has reversed direction and has a negative velocity.\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168327049595\" class=\"bc-section section\">\r\n<h3>Speed<\/h3>\r\n<p id=\"fs-id1168329146483\">In everyday language, most people use the terms <em>speed<\/em> and <em>velocity<\/em> interchangeably. In physics, however, they do not have the same meaning and are distinct concepts. One major difference is that speed has no direction; that is, speed is a scalar.<\/p>\r\n<p id=\"fs-id1168326794515\">We can calculate the <strong>average speed<\/strong> by finding the total distance traveled divided by the elapsed time:<\/p>\r\n\r\n<div id=\"fs-id1168329177306\" class=\"equation-callout\">\r\n<div id=\"fs-id1168329030544\">$$\\text{Average speed}=\\overset{\\text{\u2013}}{s}=\\frac{\\text{Total distance}}{\\text{Elapsed time}}.$$<\/div>\r\n<\/div>\r\n<p id=\"fs-id1168329155451\">Average speed is not necessarily the same as the magnitude of the average velocity, which is found by dividing the magnitude of the total displacement by the elapsed time. For example, if a trip starts and ends at the same location, the total displacement is zero, and therefore the average velocity is zero. The average speed, however, is not zero, because the total distance traveled is greater than zero. If we take a road trip of 300 km and need to be at our destination at a certain time, then we would be interested in our average speed.<\/p>\r\n<p id=\"fs-id1168329265923\">However, we can calculate the <strong>instantaneous speed<\/strong> from the magnitude of the instantaneous velocity:<\/p>\r\n\r\n<div id=\"fs-id1168326820899\" class=\"equation-callout\">\r\n<div id=\"fs-id1168329020268\">$$\\text{Instantaneous speed}=|v(t)|.$$<\/div>\r\n<\/div>\r\n<p id=\"fs-id1168329336688\">If a particle is moving along the <em>x<\/em>-axis at +7.0 m\/s and another particle is moving along the same axis at \u22127.0 m\/s, they have different velocities, but both have the same speed of 7.0 m\/s. Some typical speeds are shown in the following table.<\/p>\r\n\r\n<table id=\"fs-id1168329070697\" summary=\"This table has three columns and twelve rows. The first row is a header row and it labels each column: \u201cSpeed\u201d, \u201cmeters per second\u201d, and \u201cmiles per hour\u201d. Under the \u201cSpeed\u201d: Continental drift; Brisk walk; Cyclist; Sprint runner; Rural speed limit; Official land speed record; Speed of sound at sea level; Space shuttle on reentry; Escape velocity of Earth; Orbital speed of Earth around the Sun; Speed of light in a vacuum; Under the \u201cmeters per second\u201d: 10 in the -7 power;1.7; 4.4; 12.2; 24.6; 341.1; 343; 7800; 11,200; 29,783; 299,792,458; Under the \u201cmiles per hour\u201d: 2 multiplied by 10 in the -7 power; 3.9; 10; 27; 56; 763; 768; 17,500; 25,000; 66,623; 670,616,629;\"><caption>Speeds of Various Objects*Escape velocity is the velocity at which an object must be launched so that it overcomes Earth\u2019s gravity and is not pulled back toward Earth.<\/caption>\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Speed<\/th>\r\n<th>m\/s<\/th>\r\n<th>mi\/h<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>Continental drift<\/td>\r\n<td>$${10}^{-7}$$<\/td>\r\n<td>$$2\\,\u00d7\\,{10}^{-7}$$<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>Brisk walk<\/td>\r\n<td>1.7<\/td>\r\n<td>3.9<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>Cyclist<\/td>\r\n<td>4.4<\/td>\r\n<td>10<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>Sprint runner<\/td>\r\n<td>12.2<\/td>\r\n<td>27<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>Rural speed limit<\/td>\r\n<td>24.6<\/td>\r\n<td>56<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>Official land speed record<\/td>\r\n<td>341.1<\/td>\r\n<td>763<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>Speed of sound at sea level<\/td>\r\n<td>343<\/td>\r\n<td>768<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>Space shuttle on reentry<\/td>\r\n<td>7800<\/td>\r\n<td>17,500<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>Escape velocity of Earth*<\/td>\r\n<td>11,200<\/td>\r\n<td>25,000<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>Orbital speed of Earth around the Sun<\/td>\r\n<td>29,783<\/td>\r\n<td>66,623<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>Speed of light in a vacuum<\/td>\r\n<td>299,792,458<\/td>\r\n<td>670,616,629<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div id=\"fs-id1168329418227\" class=\"bc-section section\">\r\n<h3>Calculating Instantaneous Velocity<\/h3>\r\n<p id=\"fs-id1168329137312\">When calculating instantaneous velocity, we need to specify the explicit form of the position function <em>x<\/em>(<em>t<\/em>). For the moment, let\u2019s use polynomials $$ x(t)=A{t}^{n}$$, because they are easily differentiated using the power rule of calculus:<\/p>\r\n\r\n<div id=\"fs-id1168326841988\" class=\"equation-callout\">\r\n<div id=\"fs-id1168329070126\">$$\\frac{dx(t)}{dt}=nA{t}^{n-1}.$$<\/div>\r\n<\/div>\r\n<p id=\"fs-id1168329319372\">The following example illustrates the use of <a class=\"autogenerated-content\" href=\"#fs-id1168329070126\">(Figure)<\/a>.<\/p>\r\n\r\n<div id=\"fs-id1168329069804\" class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\n<h4>Instantaneous Velocity Versus Average Velocity<\/h4>\r\nThe position of a particle is given by $$ x(t)=3.0t+0.5{t}^{3}\\,\\text{m}$$.\r\n<ol id=\"fs-id1168326791675\" type=\"a\">\r\n \t<li>Using <a class=\"autogenerated-content\" href=\"#fs-id1168329153584\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#fs-id1168329070126\">(Figure)<\/a>, find the instantaneous velocity at $$ t=2.0 $$ s.<\/li>\r\n \t<li>Calculate the average velocity between 1.0 s and 3.0 s.<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1168329150787\">Strategy<a class=\"autogenerated-content\" href=\"#fs-id1168329153584\">(Figure)<\/a> gives the instantaneous velocity of the particle as the derivative of the position function. Looking at the form of the position function given, we see that it is a polynomial in <em>t<\/em>. Therefore, we can use <a class=\"autogenerated-content\" href=\"#fs-id1168329070126\">(Figure)<\/a>, the power rule from calculus, to find the solution. We use <a class=\"autogenerated-content\" href=\"#fs-id1168329020268\">(Figure)<\/a> to calculate the average velocity of the particle.<\/p>\r\n\r\n<h4>Solution<\/h4>\r\n<ol id=\"fs-id1168329140398\" type=\"a\">\r\n \t<li>$$v(t)=\\frac{dx(t)}{dt}=3.0+1.5{t}^{2}\\,\\text{m\/s}$$.Substituting <em>t<\/em> = 2.0 s into this equation gives $$ v(2.0\\,\\text{s})=[3.0+1.5{(2.0)}^{2}]\\,\\text{m\/s}=9.0\\,\\text{m\/s}$$.<\/li>\r\n \t<li>To determine the average velocity of the particle between 1.0 s and 3.0 s, we calculate the values of <em>x<\/em>(1.0 s) and <em>x<\/em>(3.0 s):\r\n<div id=\"fs-id1168329503288\" class=\"unnumbered\">$$x(1.0\\,\\text{s})=[(3.0)(1.0)+0.5{(1.0)}^{3}]\\,\\text{m}=3.5\\,\\text{m}$$<\/div>\r\n<div id=\"fs-id1168329266138\" class=\"unnumbered\">$$x(3.0\\,\\text{s})=[(3.0)(3.0)+0.5{(3.0)}^{3}]\\,\\text{m}=22.5\\,\\text{m.}$$<\/div>\r\nThen the average velocity is\r\n<div id=\"fs-id1168329338952\" class=\"unnumbered\">$$\\overset{\\text{\u2013}}{v}=\\frac{x(3.0\\,\\text{s})-x(1.0\\,\\text{s})}{t(3.0\\,\\text{s})-t(1.0\\,\\text{s})}=\\frac{22.5-3.5\\,\\text{m}}{3.0-1.0\\,\\text{s}}=9.5\\,\\text{m\/s}\\text{.}$$<\/div><\/li>\r\n<\/ol>\r\n<h4>Significance<\/h4>\r\nIn the limit that the time interval used to calculate $$ \\overset{\\text{\u2212}}{v} $$ goes to zero, the value obtained for $$ \\overset{\\text{\u2212}}{v} $$ converges to the value of <em>v.<\/em>\r\n\r\n<\/div>\r\n<div id=\"fs-id1168329085430\" class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\n<h4>Instantaneous Velocity Versus Speed<\/h4>\r\nConsider the motion of a particle in which the position is $$ x(t)=3.0t-3{t}^{2}\\,\\text{m}$$.\r\n<ol id=\"fs-id1168329022158\" type=\"a\">\r\n \t<li>What is the instantaneous velocity at <em>t<\/em> = 0.25 s, <em>t<\/em> = 0.50 s, and <em>t<\/em> = 1.0 s?<\/li>\r\n \t<li>What is the speed of the particle at these times?<\/li>\r\n<\/ol>\r\n<h4>Strategy<\/h4>\r\nThe instantaneous velocity is the derivative of the position function and the speed is the magnitude of the instantaneous velocity. We use <a class=\"autogenerated-content\" href=\"#fs-id1168329153584\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#fs-id1168329070126\">(Figure)<\/a> to solve for instantaneous velocity.\r\n<h4>Solution<\/h4>\r\n<ol id=\"fs-id1168326763228\" type=\"a\">\r\n \t<li>[reveal-answer q=\"225905\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"225905\"]$$v(t)=\\frac{dx(t)}{dt}=3.0-6.0t\\,\\text{m\/s}$$[\/hidden-answer]<\/li>\r\n \t<li>[reveal-answer q=\"700890\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"700890\"]$$v(0.25\\,\\text{s})=1.50\\,\\text{m\/s,}v(0.5\\,\\text{s})=0\\,\\text{m\/s,}v(1.0\\,\\text{s})=-3.0\\,\\text{m\/s}$$[\/hidden-answer]<\/li>\r\n \t<li>[reveal-answer q=\"490574\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"490574\"]$$\\text{Speed}=|v(t)|=1.50\\,\\text{m\/s},0.0\\,\\text{m\/s,}\\,\\text{and}\\,3.0\\,\\text{m\/s}$$[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<h4>Significance<\/h4>\r\nThe velocity of the particle gives us direction information, indicating the particle is moving to the left (west) or right (east). The speed gives the magnitude of the velocity. By graphing the position, velocity, and speed as functions of time, we can understand these concepts visually <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_03_02_PosVelSp\">(Figure)<\/a>. In (a), the graph shows the particle moving in the positive direction until <em>t<\/em> = 0.5 s, when it reverses direction. The reversal of direction can also be seen in (b) at 0.5 s where the velocity is zero and then turns negative. At 1.0 s it is back at the origin where it started. The particle\u2019s velocity at 1.0 s in (b) is negative, because it is traveling in the negative direction. But in (c), however, its speed is positive and remains positive throughout the travel time. We can also interpret velocity as the slope of the position-versus-time graph. The slope of <em>x<\/em>(<em>t<\/em>) is decreasing toward zero, becoming zero at 0.5 s and increasingly negative thereafter. This analysis of comparing the graphs of position, velocity, and speed helps catch errors in calculations. The graphs must be consistent with each other and help interpret the calculations.\r\n<div id=\"CNX_UPhysics_03_02_PosVelSp\" class=\"wp-caption aligncenter\">[caption id=\"\" align=\"aligncenter\" width=\"897\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184942\/CNX_UPhysics_03_02_PosVelSp.jpg\" alt=\"Graph A shows position in meters plotted versus time in seconds. It starts at the origin, reaches maximum at 0.5 seconds, and then start to decrease crossing x axis at 1 second. Graph B shows velocity in meters per second plotted as a function of time at seconds. Velocity linearly decreases from the left to the right. Graph C shows absolute velocity in meters per second plotted as a function of time at seconds. Graph has a V-leeter shape. Velocity decreases till 0.5 seconds; then it starts to increase.\" width=\"897\" height=\"274\" \/> <strong>Figure 3.9<\/strong> (a) Position: x(t) versus time. (b) Velocity: v(t) versus time. The slope of the position graph is the velocity. A rough comparison of the slopes of the tangent lines in (a) at 0.25 s, 0.5 s, and 1.0 s with the values for velocity at the corresponding times indicates they are the same values. (c) Speed: $$ |v(t)| $$ versus time. Speed is always a positive number.[\/caption]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168326919242\" class=\"textbox exercises check-understanding\">\r\n<h3>Check Your Understanding<\/h3>\r\n<div id=\"fs-id1168329179170\" class=\"problem textbox\">\r\n<div id=\"fs-id1168326791759\">\r\n<p id=\"fs-id1168329124574\">The position of an object as a function of time is $$ x(t)=-3{t}^{2}\\,\\text{m}$$. (a) What is the velocity of the object as a function of time? (b) Is the velocity ever positive? (c) What are the velocity and speed at <em>t<\/em> = 1.0 s?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1168326791024\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1168326791024\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1168326791024\"]\r\n<p id=\"fs-id1168329105566\">(a) Taking the derivative of <em>x<\/em>(<em>t<\/em>) gives <em>v<\/em>(<em>t<\/em>) = \u22126<em>t<\/em> m\/s. (b) No, because time can never be negative. (c) The velocity is <em>v<\/em>(1.0 s) = \u22126 m\/s and the speed is $$ |v(1.0\\,\\text{s})|=6\\,\\text{m\/s}$$.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168329151013\" class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"fs-id1168326904770\">\r\n \t<li>Instantaneous velocity is a continuous function of time and gives the velocity at any point in time during a particle\u2019s motion. We can calculate the instantaneous velocity at a specific time by taking the derivative of the position function, which gives us the functional form of instantaneous velocity <em>v<\/em>(<em>t<\/em>).<\/li>\r\n \t<li>Instantaneous velocity is a vector and can be negative.<\/li>\r\n \t<li>Instantaneous speed is found by taking the absolute value of instantaneous velocity, and it is always positive.<\/li>\r\n \t<li>Average speed is total distance traveled divided by elapsed time.<\/li>\r\n \t<li>The slope of a position-versus-time graph at a specific time gives instantaneous velocity at that time.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1168326822676\" class=\"review-conceptual-questions\">\r\n<h3>Conceptual Questions<\/h3>\r\n<div id=\"fs-id1168326759364\" class=\"problem textbox\">\r\n<div id=\"fs-id1168329072226\">\r\n<p id=\"fs-id1168326840996\">There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the difference between these two quantities.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1168329402518\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1168329402518\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1168329402518\"]\r\n<p id=\"fs-id1168326767452\">Average speed is the total distance traveled divided by the elapsed time. If you go for a walk, leaving and returning to your home, your average speed is a positive number. Since Average velocity = Displacement\/Elapsed time, your average velocity is zero.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168327094731\" class=\"problem textbox\">\r\n<div id=\"fs-id1168326926896\">\r\n<p id=\"fs-id1168326926898\">Does the speedometer of a car measure speed or velocity?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168326896853\" class=\"problem textbox\">\r\n<div id=\"fs-id1168326896855\">\r\n<p id=\"fs-id1168326927106\">If you divide the total distance traveled on a car trip (as determined by the odometer) by the elapsed time of the trip, are you calculating average speed or magnitude of average velocity? Under what circumstances are these two quantities the same?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1168326927108\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1168326927108\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1168326927108\"]\r\n<p id=\"fs-id1168329151650\">Average speed. They are the same if the car doesn\u2019t reverse direction.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168329340393\" class=\"problem textbox\">\r\n<div id=\"fs-id1168329340395\">\r\n<p id=\"fs-id1168327052762\">How are instantaneous velocity and instantaneous speed related to one another? How do they differ?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168326857991\" class=\"review-problems textbox exercises\">\r\n<h3>Problems<\/h3>\r\n<div id=\"fs-id1168326798408\" class=\"problem textbox\">\r\n<div id=\"fs-id1168326798410\">\r\n<p id=\"fs-id1168329466670\">A woodchuck runs 20 m to the right in 5 s, then turns and runs 10 m to the left in 3 s. (a) What is the average velocity of the woodchuck? (b) What is its average speed?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168326790679\" class=\"problem textbox\">\r\n<div id=\"fs-id1168326801116\">\r\n<p id=\"fs-id1168326801118\">Sketch the velocity-versus-time graph from the following position-versus-time graph.<\/p>\r\n<span id=\"fs-id1168326859739\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184945\/CNX_UPhysics_03_02_Prob6_img.jpg\" alt=\"Graph shows position in meters plotted versus time in seconds. It starts at the origin, reaches 4 meters at 0.4 seconds; decreases to -2 meters at 0.6 seconds, reaches minimum of -6 meters at 1 second, increases to -4 meters at 1.6 seconds, and reaches 2 meters at 2 seconds.\" \/><\/span>\r\n\r\n<\/div>\r\n<div id=\"fs-id1168326752308\"><span id=\"fs-id1168326752311\">\r\n[reveal-answer q=\"219361\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"219361\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184948\/CNX_UPhysics_03_02_Prob6ans_img.jpg\" alt=\"Graph shows velocity in meters per second plotted as a function of time at seconds. Velocity starts as 10 meters per second, decreases to -30 at 0.4 seconds; increases to -10 meters at 0.6 seconds, increases to 5 at 1 second, increases to 15 at 1.6 seconds.\" \/>[\/hidden-answer]<\/span><\/div>\r\n<\/div>\r\n<div id=\"fs-id1168328977772\" class=\"problem textbox\">\r\n<div id=\"fs-id1168326777148\">\r\n<p id=\"fs-id1168326777150\">Sketch the velocity-versus-time graph from the following position-versus-time graph.<\/p>\r\n<span id=\"fs-id1168329282325\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184950\/CNX_UPhysics_03_02_Prob7_img.jpg\" alt=\"Graph shows position plotted versus time in seconds. Graph has a sinusoidal shape. It starts with the positive value at zero time, changes to negative, and then starts to increase.\" \/><\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168329191384\" class=\"problem textbox\">\r\n<div id=\"fs-id1168326791352\">\r\n<p id=\"fs-id1168326791354\">Given the following velocity-versus-time graph, sketch the position-versus-time graph.<\/p>\r\n<span id=\"fs-id1168329503346\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184952\/CNX_UPhysics_03_02_Prob8_img.jpg\" alt=\"Graph shows velocity plotted versus time. It starts with the positive value at zero time, decreases to the negative value and remains constant.\" \/><\/span>\r\n\r\n<\/div>\r\n<div id=\"fs-id1168329187573\"><span id=\"fs-id1168329186639\">\r\n[reveal-answer q=\"284226\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"284226\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184955\/CNX_UPhysics_03_02_Prob8ans_img.jpg\" alt=\"Graph shows position plotted versus time. It starts at the origin, increases reaching maximum, and then decreases close to zero.\" \/>[\/hidden-answer]<\/span><\/div>\r\n<\/div>\r\n<div id=\"fs-id1168326762812\" class=\"problem textbox\">\r\n<div id=\"fs-id1168326762814\">\r\n<p id=\"fs-id1168326777140\">An object has a position function <em>x<\/em>(<em>t<\/em>) = 5<em>t<\/em> m. (a) What is the velocity as a function of time? (b) Graph the position function and the velocity function.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"problem textbox\">\r\n<div id=\"fs-id1168326809072\">\r\n<p id=\"fs-id1168326766071\">A particle moves along the <em>x-<\/em>axis according to $$ x(t)=10t-2{t}^{2}\\,\\text{m}$$. (a) What is the instantaneous velocity at <em>t<\/em> = 2 s and <em>t<\/em> = 3 s? (b) What is the instantaneous speed at these times? (c) What is the average velocity between <em>t<\/em> = 2 s and <em>t<\/em> = 3 s?<\/p>\r\n\r\n<\/div>\r\n<div class=\"solution\">\r\n<p id=\"fs-id1168329445296\">[reveal-answer q=\"7006\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"7006\"]a. $$ v(t)=(10-4t)\\text{m\/s}$$; v(2 s) = 2 m\/s, v(3 s) = \u22122 m\/s; b. $$ |v(2\\,\\text{s})|=2\\,\\text{m\/s},|v(3\\,\\text{s})|=2\\,\\text{m\/s}$$; (c) $$ \\overset{\\text{\u2013}}{v}=0\\,\\text{m\/s}$$[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168329265497\" class=\"problem textbox\">\r\n<div id=\"fs-id1168329265499\">\r\n<p id=\"fs-id1168329439612\"><strong>Unreasonable results.<\/strong> A particle moves along the <em>x<\/em>-axis according to $$ x(t)=3{t}^{3}+5t\\text{\u200b}$$. At what time is the velocity of the particle equal to zero? Is this reasonable?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Glossary<\/h3>\r\n<dl id=\"fs-id1168329197758\">\r\n \t<dt><strong>instantaneous velocity<\/strong><\/dt>\r\n \t<dd>the velocity at a specific instant or time point<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1168329418230\">\r\n \t<dt><strong>instantaneous speed<\/strong><\/dt>\r\n \t<dd id=\"fs-id1168326764878\">the absolute value of the instantaneous velocity<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1168329478735\">\r\n \t<dt><strong>average speed<\/strong><\/dt>\r\n \t<dd id=\"fs-id1168326838470\">the total distance traveled divided by elapsed time<\/dd>\r\n<\/dl>\r\n<\/div>","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<ul>\n<li>Explain the difference between average velocity and instantaneous velocity.<\/li>\n<li>Describe the difference between velocity and speed.<\/li>\n<li>Calculate the instantaneous velocity given the mathematical equation for the velocity.<\/li>\n<li>Calculate the speed given the instantaneous velocity.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1168326781535\">We have now seen how to calculate the average velocity between two positions. However, since objects in the real world move continuously through space and time, we would like to find the velocity of an object at any single point. We can find the velocity of the object anywhere along its path by using some fundamental principles of calculus. This section gives us better insight into the physics of motion and will be useful in later chapters.<\/p>\n<div id=\"fs-id1168329168511\" class=\"bc-section section\">\n<h3>Instantaneous Velocity<\/h3>\n<p id=\"fs-id1168329339669\">The quantity that tells us how fast an object is moving anywhere along its path is the<strong> instantaneous velocity<\/strong>, usually called simply <em>velocity<\/em>. It is the average velocity between two points on the path in the limit that the time (and therefore the displacement) between the two points approaches zero. To illustrate this idea mathematically, we need to express position <em>x<\/em> as a continuous function of <em>t<\/em> denoted by <em>x<\/em>(<em>t<\/em>). The expression for the average velocity between two points using this notation is $$ \\overset{\\text{\u2013}}{v}=\\frac{x({t}_{2})-x({t}_{1})}{{t}_{2}-{t}_{1}}$$. To find the instantaneous velocity at any position, we let $$ {t}_{1}=t $$ and $$ {t}_{2}=t+\\text{\u0394}t$$. After inserting these expressions into the equation for the average velocity and taking the limit as $$ \\text{\u0394}t\\to 0$$, we find the expression for the instantaneous velocity:<\/p>\n<div id=\"fs-id1168329127174\" class=\"unnumbered\">$$v(t)=\\underset{\\text{\u0394}t\\to 0}{\\text{lim}}\\frac{x(t+\\text{\u0394}t)-x(t)}{\\text{\u0394}t}=\\frac{dx(t)}{dt}.$$<\/div>\n<div id=\"fs-id1168329192792\">\n<div>Instantaneous Velocity<\/div>\n<p id=\"fs-id1168326802131\">The instantaneous velocity of an object is the limit of the average velocity as the elapsed time approaches zero, or the derivative of <em>x<\/em> with respect to <em>t<\/em>:<\/p>\n<div id=\"fs-id1168329153584\">$$v(t)=\\frac{d}{dt}x(t).$$<\/div>\n<\/div>\n<p id=\"fs-id1168326795408\">Like average velocity, instantaneous velocity is a vector with dimension of length per time. The instantaneous velocity at a specific time point $$ {t}_{0} $$ is the rate of change of the position function, which is the slope of the position function $$ x(t) $$ at $$ {t}_{0}$$. <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_03_02_InstVel\">(Figure)<\/a> shows how the average velocity $$ \\overset{\\text{\u2013}}{v}=\\frac{\\text{\u0394}x}{\\text{\u0394}t} $$ between two times approaches the instantaneous velocity at $$ {t}_{0}. $$ The instantaneous velocity is shown at time $$ {t}_{0}$$, which happens to be at the maximum of the position function. The slope of the position graph is zero at this point, and thus the instantaneous velocity is zero. At other times, $$ {t}_{1},{t}_{2}$$, and so on, the instantaneous velocity is not zero because the slope of the position graph would be positive or negative. If the position function had a minimum, the slope of the position graph would also be zero, giving an instantaneous velocity of zero there as well. Thus, the zeros of the velocity function give the minimum and maximum of the position function.<\/p>\n<div id=\"CNX_UPhysics_03_02_InstVel\" class=\"wp-caption aligncenter\">\n<div style=\"width: 424px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184932\/CNX_UPhysics_03_02_InstVel.jpg\" alt=\"Graph shows position plotted versus time. Position increases from t1 to t2 and reaches maximum at t0. It decreases to at and continues to decrease at t4. The slope of the tangent line at t0 is indicated as the instantaneous velocity.\" width=\"414\" height=\"352\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 3.6<\/strong> In a graph of position versus time, the instantaneous velocity is the slope of the tangent line at a given point. The average velocities $$ \\overset{\\text{\u2013}}{v}=\\frac{\\text{\u0394}x}{\\text{\u0394}t}=\\frac{{x}_{\\text{f}}-{x}_{\\text{i}}}{{t}_{\\text{f}}-{t}_{\\text{i}}} $$ between times $$ \\text{\u0394}t={t}_{6}-{t}_{1},\\text{\u0394}t={t}_{5}-{t}_{2},\\text{and}\\,\\text{\u0394}t={t}_{4}-{t}_{3} $$ are shown. When $$ \\text{\u0394}t\\to 0$$, the average velocity approaches the instantaneous velocity at $$ t={t}_{0}$$.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1168329165028\" class=\"textbox examples\">\n<h3>Example<\/h3>\n<h4>Finding Velocity from a Position-Versus-Time Graph<\/h4>\n<p>Given the position-versus-time graph of <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_03_02_PosGraph\">(Figure)<\/a>, find the velocity-versus-time graph.<\/p>\n<div id=\"CNX_UPhysics_03_02_PosGraph\" class=\"wp-caption aligncenter\">\n<div style=\"width: 456px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184935\/CNX_UPhysics_03_02_PosGraph.jpg\" alt=\"Graph shows position in kilometers plotted as a function of time at minutes. It starts at the origin, reaches 0.5 kilometers at 0.5 minutes, remains constant between 0.5 and 0.9 minutes, and decreases to 0 at 2.0 minutes.\" width=\"446\" height=\"421\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 3.7<\/strong> The object starts out in the positive direction, stops for a short time, and then reverses direction, heading back toward the origin. Notice that the object comes to rest instantaneously, which would require an infinite force. Thus, the graph is an approximation of motion in the real world. (The concept of force is discussed in Newton\u2019s Laws of Motion.)<\/p>\n<\/div>\n<\/div>\n<h4>Strategy<\/h4>\n<p>The graph contains three straight lines during three time intervals. We find the velocity during each time interval by taking the slope of the line using the grid.<\/p>\n<h4>Solution<\/h4>\n<p id=\"fs-id1168329296202\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q508307\">Show Answer<\/span><\/p>\n<div id=\"q508307\" class=\"hidden-answer\" style=\"display: none\">Time interval 0 s to 0.5 s: $$ \\overset{\\text{\u2013}}{v}=\\frac{\\text{\u0394}x}{\\text{\u0394}t}=\\frac{0.5\\,\\text{m}-0.0\\,\\text{m}}{0.5\\,\\text{s}-0.0\\,\\text{s}}=1.0\\,\\text{m\/s}$$<\/p>\n<p>Time interval 0.5 s to 1.0 s: $$ \\overset{\\text{\u2013}}{v}=\\frac{\\text{\u0394}x}{\\text{\u0394}t}=\\frac{0.0\\,\\text{m}-0.0\\,\\text{m}}{1.0\\,\\text{s}-0.5\\,\\text{s}}=0.0\\,\\text{m\/s}$$<\/p>\n<p>Time interval 1.0 s to 2.0 s: $$ \\overset{\\text{\u2013}}{v}=\\frac{\\text{\u0394}x}{\\text{\u0394}t}=\\frac{0.0\\,\\text{m}-0.5\\,\\text{m}}{2.0\\,\\text{s}-1.0\\,\\text{s}}=-0.5\\,\\text{m\/s}$$<\/p>\n<p>The graph of these values of velocity versus time is shown in (Figure).<\/p>\n<div id=\"CNX_UPhysics_03_02_VelGraph\" class=\"wp-caption aligncenter\">\n<div style=\"width: 424px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184938\/CNX_UPhysics_03_02_VelGraph.jpg\" alt=\"Graph shows velocity in meters per second plotted as a function of time at seconds. The velocity is 1 meter per second between 0 and 0.5 seconds, zero between 0.5 and 1.0 seconds, and -0.5 between 1.0 and 2.0 seconds.\" width=\"414\" height=\"315\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 3.8<\/strong> The velocity is positive for the first part of the trip, zero when the object is stopped, and negative when the object reverses direction.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h4>Significance<\/h4>\n<p>During the time interval between 0 s and 0.5 s, the object\u2019s position is moving away from the origin and the position-versus-time curve has a positive slope. At any point along the curve during this time interval, we can find the instantaneous velocity by taking its slope, which is +1 m\/s, as shown in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_03_02_VelGraph\">(Figure)<\/a>. In the subsequent time interval, between 0.5 s and 1.0 s, the position doesn\u2019t change and we see the slope is zero. From 1.0 s to 2.0 s, the object is moving back toward the origin and the slope is \u22120.5 m\/s. The object has reversed direction and has a negative velocity.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1168327049595\" class=\"bc-section section\">\n<h3>Speed<\/h3>\n<p id=\"fs-id1168329146483\">In everyday language, most people use the terms <em>speed<\/em> and <em>velocity<\/em> interchangeably. In physics, however, they do not have the same meaning and are distinct concepts. One major difference is that speed has no direction; that is, speed is a scalar.<\/p>\n<p id=\"fs-id1168326794515\">We can calculate the <strong>average speed<\/strong> by finding the total distance traveled divided by the elapsed time:<\/p>\n<div id=\"fs-id1168329177306\" class=\"equation-callout\">\n<div id=\"fs-id1168329030544\">$$\\text{Average speed}=\\overset{\\text{\u2013}}{s}=\\frac{\\text{Total distance}}{\\text{Elapsed time}}.$$<\/div>\n<\/div>\n<p id=\"fs-id1168329155451\">Average speed is not necessarily the same as the magnitude of the average velocity, which is found by dividing the magnitude of the total displacement by the elapsed time. For example, if a trip starts and ends at the same location, the total displacement is zero, and therefore the average velocity is zero. The average speed, however, is not zero, because the total distance traveled is greater than zero. If we take a road trip of 300 km and need to be at our destination at a certain time, then we would be interested in our average speed.<\/p>\n<p id=\"fs-id1168329265923\">However, we can calculate the <strong>instantaneous speed<\/strong> from the magnitude of the instantaneous velocity:<\/p>\n<div id=\"fs-id1168326820899\" class=\"equation-callout\">\n<div id=\"fs-id1168329020268\">$$\\text{Instantaneous speed}=|v(t)|.$$<\/div>\n<\/div>\n<p id=\"fs-id1168329336688\">If a particle is moving along the <em>x<\/em>-axis at +7.0 m\/s and another particle is moving along the same axis at \u22127.0 m\/s, they have different velocities, but both have the same speed of 7.0 m\/s. Some typical speeds are shown in the following table.<\/p>\n<table id=\"fs-id1168329070697\" summary=\"This table has three columns and twelve rows. The first row is a header row and it labels each column: \u201cSpeed\u201d, \u201cmeters per second\u201d, and \u201cmiles per hour\u201d. Under the \u201cSpeed\u201d: Continental drift; Brisk walk; Cyclist; Sprint runner; Rural speed limit; Official land speed record; Speed of sound at sea level; Space shuttle on reentry; Escape velocity of Earth; Orbital speed of Earth around the Sun; Speed of light in a vacuum; Under the \u201cmeters per second\u201d: 10 in the -7 power;1.7; 4.4; 12.2; 24.6; 341.1; 343; 7800; 11,200; 29,783; 299,792,458; Under the \u201cmiles per hour\u201d: 2 multiplied by 10 in the -7 power; 3.9; 10; 27; 56; 763; 768; 17,500; 25,000; 66,623; 670,616,629;\">\n<caption>Speeds of Various Objects*Escape velocity is the velocity at which an object must be launched so that it overcomes Earth\u2019s gravity and is not pulled back toward Earth.<\/caption>\n<thead>\n<tr valign=\"top\">\n<th>Speed<\/th>\n<th>m\/s<\/th>\n<th>mi\/h<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>Continental drift<\/td>\n<td>$${10}^{-7}$$<\/td>\n<td>$$2\\,\u00d7\\,{10}^{-7}$$<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>Brisk walk<\/td>\n<td>1.7<\/td>\n<td>3.9<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>Cyclist<\/td>\n<td>4.4<\/td>\n<td>10<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>Sprint runner<\/td>\n<td>12.2<\/td>\n<td>27<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>Rural speed limit<\/td>\n<td>24.6<\/td>\n<td>56<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>Official land speed record<\/td>\n<td>341.1<\/td>\n<td>763<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>Speed of sound at sea level<\/td>\n<td>343<\/td>\n<td>768<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>Space shuttle on reentry<\/td>\n<td>7800<\/td>\n<td>17,500<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>Escape velocity of Earth*<\/td>\n<td>11,200<\/td>\n<td>25,000<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>Orbital speed of Earth around the Sun<\/td>\n<td>29,783<\/td>\n<td>66,623<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>Speed of light in a vacuum<\/td>\n<td>299,792,458<\/td>\n<td>670,616,629<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id1168329418227\" class=\"bc-section section\">\n<h3>Calculating Instantaneous Velocity<\/h3>\n<p id=\"fs-id1168329137312\">When calculating instantaneous velocity, we need to specify the explicit form of the position function <em>x<\/em>(<em>t<\/em>). For the moment, let\u2019s use polynomials $$ x(t)=A{t}^{n}$$, because they are easily differentiated using the power rule of calculus:<\/p>\n<div id=\"fs-id1168326841988\" class=\"equation-callout\">\n<div id=\"fs-id1168329070126\">$$\\frac{dx(t)}{dt}=nA{t}^{n-1}.$$<\/div>\n<\/div>\n<p id=\"fs-id1168329319372\">The following example illustrates the use of <a class=\"autogenerated-content\" href=\"#fs-id1168329070126\">(Figure)<\/a>.<\/p>\n<div id=\"fs-id1168329069804\" class=\"textbox examples\">\n<h3>Example<\/h3>\n<h4>Instantaneous Velocity Versus Average Velocity<\/h4>\n<p>The position of a particle is given by $$ x(t)=3.0t+0.5{t}^{3}\\,\\text{m}$$.<\/p>\n<ol id=\"fs-id1168326791675\" type=\"a\">\n<li>Using <a class=\"autogenerated-content\" href=\"#fs-id1168329153584\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#fs-id1168329070126\">(Figure)<\/a>, find the instantaneous velocity at $$ t=2.0 $$ s.<\/li>\n<li>Calculate the average velocity between 1.0 s and 3.0 s.<\/li>\n<\/ol>\n<p id=\"fs-id1168329150787\">Strategy<a class=\"autogenerated-content\" href=\"#fs-id1168329153584\">(Figure)<\/a> gives the instantaneous velocity of the particle as the derivative of the position function. Looking at the form of the position function given, we see that it is a polynomial in <em>t<\/em>. Therefore, we can use <a class=\"autogenerated-content\" href=\"#fs-id1168329070126\">(Figure)<\/a>, the power rule from calculus, to find the solution. We use <a class=\"autogenerated-content\" href=\"#fs-id1168329020268\">(Figure)<\/a> to calculate the average velocity of the particle.<\/p>\n<h4>Solution<\/h4>\n<ol id=\"fs-id1168329140398\" type=\"a\">\n<li>$$v(t)=\\frac{dx(t)}{dt}=3.0+1.5{t}^{2}\\,\\text{m\/s}$$.Substituting <em>t<\/em> = 2.0 s into this equation gives $$ v(2.0\\,\\text{s})=[3.0+1.5{(2.0)}^{2}]\\,\\text{m\/s}=9.0\\,\\text{m\/s}$$.<\/li>\n<li>To determine the average velocity of the particle between 1.0 s and 3.0 s, we calculate the values of <em>x<\/em>(1.0 s) and <em>x<\/em>(3.0 s):\n<div id=\"fs-id1168329503288\" class=\"unnumbered\">$$x(1.0\\,\\text{s})=[(3.0)(1.0)+0.5{(1.0)}^{3}]\\,\\text{m}=3.5\\,\\text{m}$$<\/div>\n<div id=\"fs-id1168329266138\" class=\"unnumbered\">$$x(3.0\\,\\text{s})=[(3.0)(3.0)+0.5{(3.0)}^{3}]\\,\\text{m}=22.5\\,\\text{m.}$$<\/div>\n<p>Then the average velocity is<\/p>\n<div id=\"fs-id1168329338952\" class=\"unnumbered\">$$\\overset{\\text{\u2013}}{v}=\\frac{x(3.0\\,\\text{s})-x(1.0\\,\\text{s})}{t(3.0\\,\\text{s})-t(1.0\\,\\text{s})}=\\frac{22.5-3.5\\,\\text{m}}{3.0-1.0\\,\\text{s}}=9.5\\,\\text{m\/s}\\text{.}$$<\/div>\n<\/li>\n<\/ol>\n<h4>Significance<\/h4>\n<p>In the limit that the time interval used to calculate $$ \\overset{\\text{\u2212}}{v} $$ goes to zero, the value obtained for $$ \\overset{\\text{\u2212}}{v} $$ converges to the value of <em>v.<\/em><\/p>\n<\/div>\n<div id=\"fs-id1168329085430\" class=\"textbox examples\">\n<h3>Example<\/h3>\n<h4>Instantaneous Velocity Versus Speed<\/h4>\n<p>Consider the motion of a particle in which the position is $$ x(t)=3.0t-3{t}^{2}\\,\\text{m}$$.<\/p>\n<ol id=\"fs-id1168329022158\" type=\"a\">\n<li>What is the instantaneous velocity at <em>t<\/em> = 0.25 s, <em>t<\/em> = 0.50 s, and <em>t<\/em> = 1.0 s?<\/li>\n<li>What is the speed of the particle at these times?<\/li>\n<\/ol>\n<h4>Strategy<\/h4>\n<p>The instantaneous velocity is the derivative of the position function and the speed is the magnitude of the instantaneous velocity. We use <a class=\"autogenerated-content\" href=\"#fs-id1168329153584\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#fs-id1168329070126\">(Figure)<\/a> to solve for instantaneous velocity.<\/p>\n<h4>Solution<\/h4>\n<ol id=\"fs-id1168326763228\" type=\"a\">\n<li>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q225905\">Show Answer<\/span><\/p>\n<div id=\"q225905\" class=\"hidden-answer\" style=\"display: none\">$$v(t)=\\frac{dx(t)}{dt}=3.0-6.0t\\,\\text{m\/s}$$<\/div>\n<\/div>\n<\/li>\n<li>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q700890\">Show Answer<\/span><\/p>\n<div id=\"q700890\" class=\"hidden-answer\" style=\"display: none\">$$v(0.25\\,\\text{s})=1.50\\,\\text{m\/s,}v(0.5\\,\\text{s})=0\\,\\text{m\/s,}v(1.0\\,\\text{s})=-3.0\\,\\text{m\/s}$$<\/div>\n<\/div>\n<\/li>\n<li>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q490574\">Show Answer<\/span><\/p>\n<div id=\"q490574\" class=\"hidden-answer\" style=\"display: none\">$$\\text{Speed}=|v(t)|=1.50\\,\\text{m\/s},0.0\\,\\text{m\/s,}\\,\\text{and}\\,3.0\\,\\text{m\/s}$$<\/div>\n<\/div>\n<\/li>\n<\/ol>\n<h4>Significance<\/h4>\n<p>The velocity of the particle gives us direction information, indicating the particle is moving to the left (west) or right (east). The speed gives the magnitude of the velocity. By graphing the position, velocity, and speed as functions of time, we can understand these concepts visually <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_03_02_PosVelSp\">(Figure)<\/a>. In (a), the graph shows the particle moving in the positive direction until <em>t<\/em> = 0.5 s, when it reverses direction. The reversal of direction can also be seen in (b) at 0.5 s where the velocity is zero and then turns negative. At 1.0 s it is back at the origin where it started. The particle\u2019s velocity at 1.0 s in (b) is negative, because it is traveling in the negative direction. But in (c), however, its speed is positive and remains positive throughout the travel time. We can also interpret velocity as the slope of the position-versus-time graph. The slope of <em>x<\/em>(<em>t<\/em>) is decreasing toward zero, becoming zero at 0.5 s and increasingly negative thereafter. This analysis of comparing the graphs of position, velocity, and speed helps catch errors in calculations. The graphs must be consistent with each other and help interpret the calculations.<\/p>\n<div id=\"CNX_UPhysics_03_02_PosVelSp\" class=\"wp-caption aligncenter\">\n<div style=\"width: 907px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184942\/CNX_UPhysics_03_02_PosVelSp.jpg\" alt=\"Graph A shows position in meters plotted versus time in seconds. It starts at the origin, reaches maximum at 0.5 seconds, and then start to decrease crossing x axis at 1 second. Graph B shows velocity in meters per second plotted as a function of time at seconds. Velocity linearly decreases from the left to the right. Graph C shows absolute velocity in meters per second plotted as a function of time at seconds. Graph has a V-leeter shape. Velocity decreases till 0.5 seconds; then it starts to increase.\" width=\"897\" height=\"274\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 3.9<\/strong> (a) Position: x(t) versus time. (b) Velocity: v(t) versus time. The slope of the position graph is the velocity. A rough comparison of the slopes of the tangent lines in (a) at 0.25 s, 0.5 s, and 1.0 s with the values for velocity at the corresponding times indicates they are the same values. (c) Speed: $$ |v(t)| $$ versus time. Speed is always a positive number.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1168326919242\" class=\"textbox exercises check-understanding\">\n<h3>Check Your Understanding<\/h3>\n<div id=\"fs-id1168329179170\" class=\"problem textbox\">\n<div id=\"fs-id1168326791759\">\n<p id=\"fs-id1168329124574\">The position of an object as a function of time is $$ x(t)=-3{t}^{2}\\,\\text{m}$$. (a) What is the velocity of the object as a function of time? (b) Is the velocity ever positive? (c) What are the velocity and speed at <em>t<\/em> = 1.0 s?<\/p>\n<\/div>\n<div id=\"fs-id1168326791024\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1168326791024\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1168326791024\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1168329105566\">(a) Taking the derivative of <em>x<\/em>(<em>t<\/em>) gives <em>v<\/em>(<em>t<\/em>) = \u22126<em>t<\/em> m\/s. (b) No, because time can never be negative. (c) The velocity is <em>v<\/em>(1.0 s) = \u22126 m\/s and the speed is $$ |v(1.0\\,\\text{s})|=6\\,\\text{m\/s}$$.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1168329151013\" class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<ul id=\"fs-id1168326904770\">\n<li>Instantaneous velocity is a continuous function of time and gives the velocity at any point in time during a particle\u2019s motion. We can calculate the instantaneous velocity at a specific time by taking the derivative of the position function, which gives us the functional form of instantaneous velocity <em>v<\/em>(<em>t<\/em>).<\/li>\n<li>Instantaneous velocity is a vector and can be negative.<\/li>\n<li>Instantaneous speed is found by taking the absolute value of instantaneous velocity, and it is always positive.<\/li>\n<li>Average speed is total distance traveled divided by elapsed time.<\/li>\n<li>The slope of a position-versus-time graph at a specific time gives instantaneous velocity at that time.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1168326822676\" class=\"review-conceptual-questions\">\n<h3>Conceptual Questions<\/h3>\n<div id=\"fs-id1168326759364\" class=\"problem textbox\">\n<div id=\"fs-id1168329072226\">\n<p id=\"fs-id1168326840996\">There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the difference between these two quantities.<\/p>\n<\/div>\n<div id=\"fs-id1168329402518\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1168329402518\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1168329402518\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1168326767452\">Average speed is the total distance traveled divided by the elapsed time. If you go for a walk, leaving and returning to your home, your average speed is a positive number. Since Average velocity = Displacement\/Elapsed time, your average velocity is zero.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1168327094731\" class=\"problem textbox\">\n<div id=\"fs-id1168326926896\">\n<p id=\"fs-id1168326926898\">Does the speedometer of a car measure speed or velocity?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1168326896853\" class=\"problem textbox\">\n<div id=\"fs-id1168326896855\">\n<p id=\"fs-id1168326927106\">If you divide the total distance traveled on a car trip (as determined by the odometer) by the elapsed time of the trip, are you calculating average speed or magnitude of average velocity? Under what circumstances are these two quantities the same?<\/p>\n<\/div>\n<div id=\"fs-id1168326927108\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1168326927108\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1168326927108\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1168329151650\">Average speed. They are the same if the car doesn\u2019t reverse direction.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1168329340393\" class=\"problem textbox\">\n<div id=\"fs-id1168329340395\">\n<p id=\"fs-id1168327052762\">How are instantaneous velocity and instantaneous speed related to one another? How do they differ?<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1168326857991\" class=\"review-problems textbox exercises\">\n<h3>Problems<\/h3>\n<div id=\"fs-id1168326798408\" class=\"problem textbox\">\n<div id=\"fs-id1168326798410\">\n<p id=\"fs-id1168329466670\">A woodchuck runs 20 m to the right in 5 s, then turns and runs 10 m to the left in 3 s. (a) What is the average velocity of the woodchuck? (b) What is its average speed?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1168326790679\" class=\"problem textbox\">\n<div id=\"fs-id1168326801116\">\n<p id=\"fs-id1168326801118\">Sketch the velocity-versus-time graph from the following position-versus-time graph.<\/p>\n<p><span id=\"fs-id1168326859739\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184945\/CNX_UPhysics_03_02_Prob6_img.jpg\" alt=\"Graph shows position in meters plotted versus time in seconds. It starts at the origin, reaches 4 meters at 0.4 seconds; decreases to -2 meters at 0.6 seconds, reaches minimum of -6 meters at 1 second, increases to -4 meters at 1.6 seconds, and reaches 2 meters at 2 seconds.\" \/><\/span><\/p>\n<\/div>\n<div id=\"fs-id1168326752308\"><span id=\"fs-id1168326752311\"><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q219361\">Show Answer<\/span><\/p>\n<div id=\"q219361\" class=\"hidden-answer\" style=\"display: none\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184948\/CNX_UPhysics_03_02_Prob6ans_img.jpg\" alt=\"Graph shows velocity in meters per second plotted as a function of time at seconds. Velocity starts as 10 meters per second, decreases to -30 at 0.4 seconds; increases to -10 meters at 0.6 seconds, increases to 5 at 1 second, increases to 15 at 1.6 seconds.\" \/><\/div>\n<\/div>\n<p><\/span><\/div>\n<\/div>\n<div id=\"fs-id1168328977772\" class=\"problem textbox\">\n<div id=\"fs-id1168326777148\">\n<p id=\"fs-id1168326777150\">Sketch the velocity-versus-time graph from the following position-versus-time graph.<\/p>\n<p><span id=\"fs-id1168329282325\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184950\/CNX_UPhysics_03_02_Prob7_img.jpg\" alt=\"Graph shows position plotted versus time in seconds. Graph has a sinusoidal shape. It starts with the positive value at zero time, changes to negative, and then starts to increase.\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1168329191384\" class=\"problem textbox\">\n<div id=\"fs-id1168326791352\">\n<p id=\"fs-id1168326791354\">Given the following velocity-versus-time graph, sketch the position-versus-time graph.<\/p>\n<p><span id=\"fs-id1168329503346\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184952\/CNX_UPhysics_03_02_Prob8_img.jpg\" alt=\"Graph shows velocity plotted versus time. It starts with the positive value at zero time, decreases to the negative value and remains constant.\" \/><\/span><\/p>\n<\/div>\n<div id=\"fs-id1168329187573\"><span id=\"fs-id1168329186639\"><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q284226\">Show Answer<\/span><\/p>\n<div id=\"q284226\" class=\"hidden-answer\" style=\"display: none\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184955\/CNX_UPhysics_03_02_Prob8ans_img.jpg\" alt=\"Graph shows position plotted versus time. It starts at the origin, increases reaching maximum, and then decreases close to zero.\" \/><\/div>\n<\/div>\n<p><\/span><\/div>\n<\/div>\n<div id=\"fs-id1168326762812\" class=\"problem textbox\">\n<div id=\"fs-id1168326762814\">\n<p id=\"fs-id1168326777140\">An object has a position function <em>x<\/em>(<em>t<\/em>) = 5<em>t<\/em> m. (a) What is the velocity as a function of time? (b) Graph the position function and the velocity function.<\/p>\n<\/div>\n<\/div>\n<div class=\"problem textbox\">\n<div id=\"fs-id1168326809072\">\n<p id=\"fs-id1168326766071\">A particle moves along the <em>x-<\/em>axis according to $$ x(t)=10t-2{t}^{2}\\,\\text{m}$$. (a) What is the instantaneous velocity at <em>t<\/em> = 2 s and <em>t<\/em> = 3 s? (b) What is the instantaneous speed at these times? (c) What is the average velocity between <em>t<\/em> = 2 s and <em>t<\/em> = 3 s?<\/p>\n<\/div>\n<div class=\"solution\">\n<p id=\"fs-id1168329445296\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q7006\">Show Answer<\/span><\/p>\n<div id=\"q7006\" class=\"hidden-answer\" style=\"display: none\">a. $$ v(t)=(10-4t)\\text{m\/s}$$; v(2 s) = 2 m\/s, v(3 s) = \u22122 m\/s; b. $$ |v(2\\,\\text{s})|=2\\,\\text{m\/s},|v(3\\,\\text{s})|=2\\,\\text{m\/s}$$; (c) $$ \\overset{\\text{\u2013}}{v}=0\\,\\text{m\/s}$$<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1168329265497\" class=\"problem textbox\">\n<div id=\"fs-id1168329265499\">\n<p id=\"fs-id1168329439612\"><strong>Unreasonable results.<\/strong> A particle moves along the <em>x<\/em>-axis according to $$ x(t)=3{t}^{3}+5t\\text{\u200b}$$. At what time is the velocity of the particle equal to zero? Is this reasonable?<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl id=\"fs-id1168329197758\">\n<dt><strong>instantaneous velocity<\/strong><\/dt>\n<dd>the velocity at a specific instant or time point<\/dd>\n<\/dl>\n<dl id=\"fs-id1168329418230\">\n<dt><strong>instantaneous speed<\/strong><\/dt>\n<dd id=\"fs-id1168326764878\">the absolute value of the instantaneous velocity<\/dd>\n<\/dl>\n<dl id=\"fs-id1168329478735\">\n<dt><strong>average speed<\/strong><\/dt>\n<dd id=\"fs-id1168326838470\">the total distance traveled divided by elapsed time<\/dd>\n<\/dl>\n<\/div>\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-353\">\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>OpenStax University Physics. <strong>Authored by<\/strong>: OpenStax CNX. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/cnx.org\/contents\/1Q9uMg_a@10.16:Gofkr9Oy@15\">https:\/\/cnx.org\/contents\/1Q9uMg_a@10.16:Gofkr9Oy@15<\/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>: Download for free at http:\/\/cnx.org\/contents\/d50f6e32-0fda-46ef-a362-9bd36ca7c97d@10.16<\/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":311,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"OpenStax University Physics\",\"author\":\"OpenStax CNX\",\"organization\":\"\",\"url\":\"https:\/\/cnx.org\/contents\/1Q9uMg_a@10.16:Gofkr9Oy@15\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/d50f6e32-0fda-46ef-a362-9bd36ca7c97d@10.16\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":"all-rights-reserved"},"chapter-type":[],"contributor":[],"license":[56],"class_list":["post-353","chapter","type-chapter","status-publish","hentry","license-all-rights-reserved"],"part":309,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters\/353","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/users\/311"}],"version-history":[{"count":10,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters\/353\/revisions"}],"predecessor-version":[{"id":2215,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters\/353\/revisions\/2215"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/parts\/309"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters\/353\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/media?parent=353"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapter-type?post=353"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/contributor?post=353"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/license?post=353"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}