{"id":342,"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=342"},"modified":"2016-11-03T18:08:22","modified_gmt":"2016-11-03T18:08:22","slug":"2-8-graphical-analysis-of-one-dimensional-motion","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/chapter\/2-8-graphical-analysis-of-one-dimensional-motion\/","title":{"raw":"Graphical Analysis of One-Dimensional Motion","rendered":"Graphical Analysis of One-Dimensional Motion"},"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\">Describe a straight-line graph in terms of its slope and <span class=\"foreignphrase\"><em class=\"foreignphrase\">y<\/em><\/span>-intercept.<\/li>\r\n \t<li class=\"listitem\">Determine average velocity or instantaneous velocity from a graph of position vs. time.<\/li>\r\n \t<li class=\"listitem\">Determine average or instantaneous acceleration from a graph of velocity vs. time.<\/li>\r\n \t<li class=\"listitem\">Derive a graph of velocity vs. time from a graph of position vs. time.<\/li>\r\n \t<li class=\"listitem\">Derive a graph of acceleration vs. time from a graph of velocity vs. time.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\nA graph, like a picture, is worth a thousand words. Graphs not only contain numerical information; they also reveal relationships between physical quantities. This section uses graphs of displacement, velocity, and acceleration versus time to illustrate one-dimensional kinematics.\r\n<div class=\"section\" title=\"Slopes and General Relationships\">\r\n<div class=\"titlepage\">\r\n<div>\r\n<div>\r\n<h2 id=\"m42103-fs-id1396690\"><span class=\"cnx-gentext-section cnx-gentext-t\">Slopes and General Relationships<\/span><\/h2>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nFirst note that graphs in this text have perpendicular axes, one horizontal and the other vertical. When two physical quantities are plotted against one another in such a graph, the horizontal axis is usually considered to be an <em class=\"glossterm\"> independent variable<\/em><a id=\"id783882\" class=\"indexterm\"><\/a> and the vertical axis a <em class=\"glossterm\"> dependent variable<\/em><a id=\"id783896\" class=\"indexterm\"><\/a>. If we call the horizontal axis the <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>-axis and the vertical axis the <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>y<\/em><\/span><\/span>-axis, as in Figure 1, a straight-line graph has the general form\r\n<div id=\"m42103-import-auto-id4175150\" class=\"equation\" title=\"Equation 2.101.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]y=\\text{mx}+b\\\\[\/latex]<\/div>\r\n<\/div>\r\nHere <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>m<\/em><\/span><\/span> is the <em class=\"glossterm\"> slope<\/em><a id=\"id784470\" class=\"indexterm\"><\/a>, defined to be the rise divided by the run (as seen in the figure) of the straight line. The letter <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>b<\/em><\/span><\/span> is used for the <em class=\"glossterm\"> <span class=\"emphasis\"><em>y<\/em><\/span>-intercept<\/em><a id=\"id784596\" class=\"indexterm\"><\/a>, which is the point at which the line crosses the vertical axis.\r\n<div id=\"m42103-import-auto-id2359358\" class=\"figure\" title=\"Figure 2.46.\">\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\/20101731\/Figure_02_07_01.jpg\" alt=\"Graph of a straight-line sloping up at about 40 degrees.\" width=\"300\" height=\"308\" \/> Figure 1. A straight-line graph. The equation for a straight line is .[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"section\" title=\"Graph of Displacement vs. Time (a = 0, so v is constant)\">\r\n<div class=\"titlepage\">\r\n<div>\r\n<div>\r\n<h2 id=\"m42103-fs-id2201114\"><span class=\"cnx-gentext-section cnx-gentext-t\">Graph of Displacement vs. Time (<span class=\"emphasis\"><em>a<\/em><\/span> = 0, so <span class=\"emphasis\"><em>v<\/em><\/span> is constant)<\/span><\/h2>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nTime is usually an independent variable that other quantities, such as displacement, depend upon. A graph of displacement versus time would, thus, have <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> on the vertical axis and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> on the horizontal axis. Figure 2\u00a0is just such a straight-line graph. It shows a graph of displacement versus time for a jet-powered car on a very flat dry lake bed in Nevada.\r\n<div id=\"m42103-import-auto-id2574769\" class=\"figure\" title=\"Figure 2.47.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"450\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101733\/Figure_02_07_02.jpg\" alt=\"Line graph of jet car displacement in meters versus time in seconds. The line is straight with a positive slope. The y intercept is four hundred meters. The total change in time is eight point zero seconds. The initial position is four hundred meters. The final position is two thousand meters.\" width=\"450\" height=\"440\" \/> Figure 2. Graph of displacement versus time for a jet-powered car on the Bonneville Salt Flats.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nUsing the relationship between dependent and independent variables, we see that the slope in the graph above is average velocity [latex]\\bar{v}\\\\[\/latex]\u00a0and the intercept is displacement at time zero\u2014that is, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>0<\/sub><\/span>. Substituting these symbols into [latex]y=\\text{mx}+b\\\\[\/latex]\u00a0gives\r\n<div id=\"m42103-import-auto-id4019047\" class=\"equation\" title=\"Equation 2.102.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]x=\\bar{v}t+{x}_{0}\\\\[\/latex]<\/div>\r\n<\/div>\r\nor\r\n<div id=\"m42103-import-auto-id2357165\" class=\"equation\" title=\"Equation 2.103.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]x={x}_{0}+\\bar{v}t\\\\[\/latex].<\/div>\r\n<\/div>\r\nThus a graph of displacement versus time gives a general relationship among displacement, velocity, and time, as well as giving detailed numerical information about a specific situation.\r\n<div id=\"m42103-fs-id4125096\" class=\"note\">\r\n<div class=\"title textbox shaded\">\r\n<h3 class=\"title\"><strong><span class=\"cnx-gentext-tip-t\">The Slope of <span class=\"emphasis\"><em>x<\/em><\/span> vs. <span class=\"emphasis\"><em>t<\/em><\/span> <\/span><\/strong><\/h3>\r\n<div class=\"body\">\r\n\r\nThe slope of the graph of displacement <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> vs. time <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> is velocity <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span>.\r\n<p style=\"text-align: center;\">[latex]\\text{slope}=\\frac{\\Delta x}{\\Delta t}=v\\\\[\/latex]<\/p>\r\nNotice that this equation is the same as that derived algebraically from other motion equations in <a class=\"link\" title=\"2.5. Motion Equations for Constant Acceleration in One Dimension\" href=\".\/chapter\/2-5-motion-equations-for-constant-acceleration-in-one-dimension\/\" target=\"_blank\">Motion Equations for Constant Acceleration in One Dimension<\/a>.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nFrom the figure we can see that the car has a displacement of 400 m at time 0.650 m at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> = 1.0 s, and so on. Its displacement at times other than those listed in the table can be read from the graph; furthermore, information about its velocity and acceleration can also be obtained from the graph.\r\n<div class=\"textbox examples\">\r\n<h3>Example 1. Determining Average Velocity from a Graph of Displacement versus Time: Jet Car<\/h3>\r\n<div class=\"body\">\r\n\r\nFind the average velocity of the car whose position is graphed in Figure 2.\r\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\r\nThe slope of a graph of <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> vs. <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> is average velocity, since slope equals rise over run. In this case, rise = change in displacement and run = change in time, so that\r\n<div id=\"m42103-import-auto-id1850584\" class=\"equation\" title=\"Equation 2.105.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]\\text{slope}=\\frac{\\Delta x}{\\Delta t}=\\bar{v}\\\\[\/latex].<\/div>\r\n<div class=\"mediaobject\"><\/div>\r\n<\/div>\r\nSince the slope is constant here, any two points on the graph can be used to find the slope. (Generally speaking, it is most accurate to use two widely separated points on the straight line. This is because any error in reading data from the graph is proportionally smaller if the interval is larger.)\r\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\r\n1. Choose two points on the line. In this case, we choose the points labeled on the graph: (6.4 s, 2000 m) and (0.50 s, 525 m). (Note, however, that you could choose any two points.)\r\n\r\n2. Substitute the <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> values of the chosen points into the equation. Remember in calculating change <span class=\"token\">(\u0394)<\/span> we always use final value minus initial value.\r\n<div id=\"m42103-import-auto-id4181834\" class=\"equation\" title=\"Equation 2.106.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]\\bar{v}=\\frac{\\Delta x}{\\Delta t}=\\frac{\\text{2000 m}-\\text{525 m}}{6\\text{.}\\text{4 s}-0\\text{.}\\text{50 s}}\\\\[\/latex],<\/div>\r\n<\/div>\r\nyielding\r\n<div id=\"m42103-import-auto-id2024433\" class=\"equation\" title=\"Equation 2.107.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]\\bar{v}=\\text{250 m\/s}[\/latex].<\/div>\r\n<div class=\"mediaobject\"><\/div>\r\n<\/div>\r\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\r\nThis is an impressively large land speed (900 km\/h, or about 560 mi\/h): much greater than the typical highway speed limit of 60 mi\/h (27 m\/s or 96 km\/h), but considerably shy of the record of 343 m\/s (1234 km\/h or 766 mi\/h) set in 1997.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"section\" title=\"Graphs of Motion when a is constant but a\u22600\">\r\n<div class=\"titlepage\">\r\n<div>\r\n<div>\r\n<h2 id=\"m42103-fs-id2570304\"><span class=\"cnx-gentext-section cnx-gentext-t\">Graphs of Motion when <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span> is constant but <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span>\u22600<\/span><\/span><\/h2>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nThe graphs in Figure 3\u00a0below represent the motion of the jet-powered car as it accelerates toward its top speed, but only during the time when its acceleration is constant. Time starts at zero for this motion (as if measured with a stopwatch), and the displacement and velocity are initially 200 m and 15 m\/s, respectively.\r\n<div id=\"m42103-import-auto-id3596921\" class=\"figure\" title=\"Figure 2.48.\">\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\/20101738\/Figure_02_07_03.jpg\" alt=\"Three line graphs. First is a line graph of displacement over time. Line has a positive slope that increases with time. Second line graph is of velocity over time. Line is straight with a positive slope. Third line graph is of acceleration over time. Line is straight and horizontal, indicating constant acceleration.\" width=\"300\" height=\"1397\" \/> Figure 3. Graphs of motion of a jet-powered car during the time span when its acceleration is constant. (a) The slope of an x vs. t graph is velocity. This is shown at two points, and the instantaneous velocities obtained are plotted in the next graph. Instantaneous velocity at any point is the slope of the tangent at that point. (b) The slope of the v vs. t graph is constant for this part of the motion, indicating constant acceleration. (c) Acceleration has the constant value of 5.0 m\/s2 over the time interval plotted.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\"><\/div>\r\n<\/div>\r\n<div id=\"m42103-import-auto-id3583460\" class=\"figure\" title=\"Figure 2.49.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"alignright\" width=\"300\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101740\/Figure_02_07_03a.jpg\" alt=\"Figure (Figure_02_07_03a.jpg)\" width=\"300\" height=\"450\" \/> Figure 4. A U.S. Air Force jet car speeds down a track. (credit: Matt Trostle, Flickr)[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\">The graph of displacement versus time in Figure\u00a03(a) is a curve rather than a straight line. The slope of the curve becomes steeper as time progresses, showing that the velocity is increasing over time. The slope at any point on a displacement-versus-time graph is the instantaneous velocity at that point. It is found by drawing a straight line tangent to the curve at the point of interest and taking the slope of this straight line. Tangent lines are shown for two points in Figure 3(a). If this is done at every point on the curve and the values are plotted against time, then the graph of velocity versus time shown in Figure 3(b) is obtained. Furthermore, the slope of the graph of velocity versus time is acceleration, which is shown in Figure 3(c).<\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 2. Determining Instantaneous Velocity from the Slope at a Point: Jet Car<\/h3>\r\n<div class=\"section\" title=\"Graphs of Motion when a is constant but a\u22600\">\r\n<div class=\"body\">\r\n\r\nCalculate the velocity of the jet car at a time of 25 s by finding the slope of the <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> vs. <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> graph in the graph below.\r\n<div id=\"m42103-import-auto-id4141386\" class=\"figure\" title=\"Figure 2.50.\">\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\/20101741\/graphics41.jpg\" alt=\"A graph of displacement versus time for a jet car. The x axis for time runs from zero to thirty five seconds. The y axis for displacement runs from zero to three thousand meters. The curve depicting displacement is concave up. The slope of the curve increases over time. Slope equals velocity v. There are two points on the curve, labeled, P and Q. P is located at time equals ten seconds. Q is located and time equals twenty-five seconds. A line tangent to P at ten seconds is drawn and has a slope delta x sub P over delta t sub p. A line tangent to Q at twenty five seconds is drawn and has a slope equal to delta x sub q over delta t sub q. Select coordinates are given in a table and consist of the following: time zero seconds displacement two hundred meters; time five seconds displacement three hundred thirty eight meters; time ten seconds displacement six hundred meters; time fifteen seconds displacement nine hundred eighty eight meters. Time twenty seconds displacement one thousand five hundred meters; time twenty five seconds displacement two thousand one hundred thirty eight meters; time thirty seconds displacement two thousand nine hundred meters.\" width=\"300\" height=\"368\" \/> Figure 5.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\"><\/div>\r\n<div class=\"caption\">The slope of an <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> vs. <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> graph is velocity. This is shown at two points. Instantaneous velocity at any point is the slope of the tangent at that point.<\/div>\r\n<div class=\"caption\"><\/div>\r\n<\/div>\r\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\r\nThe slope of a curve at a point is equal to the slope of a straight line tangent to the curve at that point. This principle is illustrated in Figure 5, where Q is the point at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span>=25 s<\/span>.\r\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\r\n1. Find the tangent line to the curve at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>= 25 s<\/span>.\r\n\r\n2. Determine the endpoints of the tangent. These correspond to a position of 1300 m at time 19 s and a position of 3120 m at time 32 s.\r\n\r\n3. Plug these endpoints into the equation to solve for the slope, <span class=\"emphasis\"><em><span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span><\/em><\/span>.\r\n<div id=\"m42103-import-auto-id3627758\" class=\"equation\" title=\"Equation 2.108.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]\\text{slope}={v}_{Q}=\\frac{{\\Delta x}_{Q}}{{\\Delta t}_{Q}}=\\frac{\\left(\\text{3120 m}-\\text{1300 m}\\right)}{\\left(\\text{32 s}-\\text{19 s}\\right)}\\\\[\/latex]<\/div>\r\n<div class=\"mediaobject\"><\/div>\r\n<\/div>\r\nThus,\r\n<p style=\"text-align: center;\">[latex]{v}_{Q}=\\frac{\\text{1820 m}}{\\text{13 s}}=\\text{140 m\/s.}\\\\[\/latex]<\/p>\r\n\r\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\r\nThis is the value given in this figure\u2019s table for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span> at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>= 25 s<\/span>. The value of 140 m\/s for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>Q<\/sub><\/span> is plotted in Figure 5. The entire graph of <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span> vs. <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> can be obtained in this fashion.\r\n\r\n<\/div>\r\n<\/div>\r\nCarrying this one step further, we note that the slope of a velocity versus time graph is acceleration. Slope is rise divided by run; on a <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span> vs. <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> graph, rise = change in velocity <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span> and run = change in time <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span>.\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"m42103-fs-id1405001\" class=\"note\">\r\n<div class=\"title textbox shaded\">\r\n<h3 class=\"title\"><strong><span class=\"cnx-gentext-tip-t\">The Slope of <span class=\"emphasis\"><em>v<\/em><\/span> vs. <span class=\"emphasis\"><em>t<\/em><\/span><\/span><\/strong><\/h3>\r\n<div class=\"body\">\r\n\r\nThe slope of a graph of velocity <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span> vs. time <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> is acceleration <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span><span class=\"emphasis\"><em>.<\/em><\/span>\r\n<div id=\"m42103-import-auto-id4096826\" class=\"equation\" title=\"Equation 2.110.\">\r\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]\\text{slope}=\\frac{\\Delta v}{\\Delta t}=a\\\\[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nSince the velocity versus time graph in Figure 3(b) is a straight line, its slope is the same everywhere, implying that acceleration is constant. Acceleration versus time is graphed in Figure 3(c).\r\n\r\nAdditional general information can be obtained from Figure 5\u00a0and the expression for a straight line,[latex]y=\\text{mx}+b\\\\[\/latex].\r\n\r\nIn this case, the vertical axis <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>y<\/em><\/span><\/span> is <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>V<\/em><\/span><\/span>, the intercept <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>b<\/em><\/span><\/span> is <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0<\/sub><\/span>, the slope <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>m<\/em><\/span><\/span> is <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span>, and the horizontal axis <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> is <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span>. Substituting these symbols yields\r\n<p style=\"text-align: center;\">[latex]v={v}_{0}+\\text{at}\\\\[\/latex].<\/p>\r\nA general relationship for velocity, acceleration, and time has again been obtained from a graph. Notice that this equation was also derived algebraically from other motion equations in <a class=\"link\" title=\"2.5. Motion Equations for Constant Acceleration in One Dimension\" href=\".\/chapter\/2-5-motion-equations-for-constant-acceleration-in-one-dimension\/\" target=\"_blank\">Motion Equations for Constant Acceleration in One Dimension<\/a>.\r\n\r\nIt is not accidental that the same equations are obtained by graphical analysis as by algebraic techniques. In fact, an important way to <span class=\"emphasis\"><em>discover<\/em><\/span> physical relationships is to measure various physical quantities and then make graphs of one quantity against another to see if they are correlated in any way. Correlations imply physical relationships and might be shown by smooth graphs such as those above. From such graphs, mathematical relationships can sometimes be postulated. Further experiments are then performed to determine the validity of the hypothesized relationships.\r\n<div class=\"section\" title=\"Graphs of Motion Where Acceleration is Not Constant\">\r\n<div class=\"titlepage\">\r\n<div>\r\n<div>\r\n<h2 id=\"m42103-fs-id2306208\"><span class=\"cnx-gentext-section cnx-gentext-t\">Graphs of Motion Where Acceleration is Not Constant<\/span><\/h2>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nNow consider the motion of the jet car as it goes from 165 m\/s to its top velocity of 250 m\/s, graphed in Figure 6. Time again starts at zero, and the initial displacement and velocity are 2900 m and 165 m\/s, respectively. (These were the final displacement and velocity of the car in the motion graphed in Figure 3.) Acceleration gradually decreases from <span class=\"token\">5.0 m\/s<sup>2<\/sup><\/span> to zero when the car hits 250 m\/s. The slope of the <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> vs. <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> graph increases until <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span>=55 s<\/span>, after which time the slope is constant. Similarly, velocity increases until 55 s and then becomes constant, since acceleration decreases to zero at 55 s and remains zero afterward.\r\n<div id=\"m42103-import-auto-id1534076\" class=\"figure\" title=\"Figure 2.51.\">\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\/20101745\/Figure_02_07_04.jpg\" alt=\"Three line graphs of jet car displacement, velocity, and acceleration, respectively. First line graph is of position over time. Line is straight with a positive slope. Second line graph is of velocity over time. Line graph has a positive slope that decreases over time and flattens out at the end. Third line graph is of acceleration over time. Line has a negative slope that increases over time until it flattens out at the end. The line is not smooth, but has several kinks.\" width=\"350\" height=\"1327\" \/> Figure 6. Graphs of motion of a jet-powered car as it reaches its top velocity. This motion begins where the motion in Figure 3 ends. (a) The slope of this graph is velocity; it is plotted in the next graph. (b) The velocity gradually approaches its top value. The slope of this graph is acceleration; it is plotted in the final graph. (c) Acceleration gradually declines to zero when velocity becomes constant.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\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\">\u00a0<\/span><span class=\"cnx-gentext-example cnx-gentext-t\">Calculating Acceleration from a Graph of Velocity versus Time<\/span><\/h3>\r\n<div id=\"m42103-import-auto-id1534076\" class=\"figure\" title=\"Figure 2.51.\">\r\n<div class=\"body\">\r\n\r\nCalculate the acceleration of the jet car at a time of 25 s by finding the slope of the <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span> vs. <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> graph in Figure 6(b).\r\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\r\nThe slope of the curve at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>= 25 s<\/span> is equal to the slope of the line tangent at that point, as illustrated in Figure 6(b).\r\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\r\nDetermine endpoints of the tangent line from the figure, and then plug them into the equation to solve for slope, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span>.\r\n<div id=\"import-auto-id3503054\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\text{slope}=\\frac{\\Delta v}{\\Delta t}=\\frac{\\left(\\text{260 m\/s}-\\text{210 m\/s}\\right)}{\\left(\\text{51 s}-1.0 s\\right)}\\\\[\/latex]<\/div>\r\n<div id=\"import-auto-id2028886\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]a=\\frac{\\text{50 m\/s}}{\\text{50 s}}=1\\text{.}0 m{\\text{\/s}}^{2}\\\\[\/latex].<\/div>\r\n<div id=\"m42103-import-auto-id2028886\" class=\"equation\" title=\"Equation 2.113.\"><\/div>\r\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\r\nNote that this value for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span> is consistent with the value plotted in Figure 6(c) at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>= 25 s<\/span>.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nA graph of displacement versus time can be used to generate a graph of velocity versus time, and a graph of velocity versus time can be used to generate a graph of acceleration versus time. We do this by finding the slope of the graphs at every point. If the graph is linear (i.e., a line with a constant slope), it is easy to find the slope at any point and you have the slope for every point. Graphical analysis of motion can be used to describe both specific and general characteristics of kinematics. Graphs can also be used for other topics in physics. An important aspect of exploring physical relationships is to graph them and look for underlying relationships.\r\n<div id=\"m42103-fs-id1571006\" class=\"exercise labeled check-understanding\">\r\n<div class=\"textbox key-takeaways\">\r\n<div id=\"m42103-fs-id1571006\" class=\"exercise labeled check-understanding\">\r\n<h3 class=\"title\">Check Your Understanding<\/h3>\r\nA graph of velocity vs. time of a ship coming into a harbor is shown below. (a) Describe the motion of the ship based on the graph. (b)What would a graph of the ship\u2019s acceleration look like?\r\n<div id=\"m42103-import-auto-id3504346\" class=\"figure\" title=\"Figure 2.51.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"200\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101750\/Figure_02_07_04a.jpg\" alt=\"Line graph of velocity versus time. The line has three legs. The first leg is flat. The second leg has a negative slope. The third leg also has a negative slope, but the slope is not as negative as the second leg.\" width=\"200\" height=\"193\" \/> Figure 7.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"m42103-fs-id1658952\" class=\"solution\">\r\n<h4 class=\"title\"><strong>Solution<\/strong><\/h4>\r\n<div class=\"title\"><\/div>\r\n<div class=\"body\">\r\n\r\n(a) The ship moves at constant velocity and then begins to decelerate at a constant rate. At some point, its deceleration rate decreases. It maintains this lower deceleration rate until it stops moving.\r\n\r\n(b) A graph of acceleration vs. time would show zero acceleration in the first leg, large and constant negative acceleration in the second leg, and constant negative acceleration.\r\n<div id=\"m42103-import-auto-id1666671\" class=\"figure\" title=\"Figure 2.51.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"200\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20101752\/Figure_02_07_04b.jpg\" alt=\"A line graph of acceleration versus time. There are three legs of the graph. All three legs are flat and straight. The first leg shows constant acceleration of 0. The second leg shows a constant negative acceleration. The third leg shows a constant negative acceleration that is not as negative as the second leg.\" width=\"200\" height=\"193\" \/> Figure 8.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\"><\/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>Graphs of motion can be used to analyze motion.<\/li>\r\n \t<li>Graphical solutions yield identical solutions to mathematical methods for deriving motion equations.<\/li>\r\n \t<li>The slope of a graph of displacement <em>x<\/em> vs. time\u00a0<em>t<\/em>\u00a0is velocity <em>v<\/em>.<\/li>\r\n \t<li>The slope of a graph of velocity <em>v<\/em> vs. <em>t<\/em> graph is acceleration<em> a<\/em>.<\/li>\r\n \t<li>Average velocity, instantaneous velocity, and acceleration can all be obtained by analyzing graphs.<\/li>\r\n<\/ul>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Conceptual Questions<\/h3>\r\n<div id=\"m42103-fs-id1571006\" class=\"exercise labeled check-understanding\">\r\n\r\n1. (a) Explain how you can use the graph of position versus time in Figure 9 to describe the change in velocity over time. Identify:\u00a0(b) the time (<span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>a<\/sub><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>b<\/sub><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>c<\/sub><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>d<\/sub><\/span>, or <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>e<\/sub><\/span>) at which the instantaneous velocity is greatest,\u00a0(c) the time at which it is zero, and\u00a0(d) the time at which it is negative.\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\/1322\/2015\/12\/03205558\/Figure_03_08Sol_01.jpg\" alt=\"Line graph of position versus time with 5 points labeled: a, b, c, d, and e. The slope of the line changes. It begins with a positive slope that decreases over time until around point d, where it is flat. It then has a slightly negative slope.\" width=\"300\" height=\"411\" data-media-type=\"image\/jpg\" \/> Figure 9.[\/caption]\r\n\r\n2. (a) Sketch a graph of velocity versus time corresponding to the graph of displacement versus time given in Figure 10.\u00a0(b) Identify the time or times (<span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>a<\/sub><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>b<\/sub><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>c<\/sub><\/span>, etc.) at which the instantaneous velocity is greatest.\u00a0(c) At which times is it zero?\u00a0(d) At which times is it negative?\r\n\r\n<\/div>\r\n<div class=\"title\"><\/div>\r\n<div class=\"body\">\r\n<div class=\"problem\">\r\n<div id=\"m42103-import-auto-id2562897\" class=\"figure\" title=\"Figure 2.51.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n<div class=\"wp-caption alignnone\">\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\/20101800\/Figure_03_08Sol_02.jpg\" alt=\"Line graph of position over time with 12 points labeled a through l. Line has a negative slope from a to c, where it turns and has a positive slope till point e. It turns again and has a negative slope till point g. The slope then increases again till l, where it flattens out.\" width=\"300\" height=\"416\" \/> Figure 10.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n3. (a) Explain how you can determine the acceleration over time from a velocity versus time graph such as the one in Figure 11.\u00a0(b) Based on the graph, how does acceleration change over time?\r\n<div class=\"body\">\r\n<div class=\"problem\">\r\n<div id=\"m42103-import-auto-id1778975\" class=\"figure\" title=\"Figure 2.51.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n<div class=\"wp-caption alignnone\">\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\/20101801\/Figure_03_08Sol_04.jpg\" alt=\"Line graph of velocity over time with two points labeled. Point P is at v 1 t 1. Point Q is at v 2 t 2. The line has a positive slope that increases over time.\" width=\"300\" height=\"391\" \/> Figure 11.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n4. \u00a0(a) Sketch a graph of acceleration versus time corresponding to the graph of velocity versus time given in Figure 12.\u00a0(b) Identify the time or times (<span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>a<\/sub><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>b<\/sub><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>c<\/sub><\/span>, etc.) at which the acceleration is greatest.\u00a0(c) At which times is it zero?\u00a0(d) At which times is it negative?\r\n<div class=\"body\">\r\n<div class=\"problem\">\r\n<div id=\"m42103-import-auto-id1447833\" class=\"figure\" title=\"Figure 2.51.\">\r\n<div class=\"body\">\r\n<div class=\"wp-caption alignnone\">\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\/20101802\/Figure_03_08Sol_05.jpg\" alt=\"Line graph of velocity over time with 12 points labeled a through l. The line has a positive slope from a at the origin to d where it slopes downward to e, and then back upward to h. It then slopes back down to point l at v equals 0.\" width=\"300\" height=\"408\" \/> Figure 12.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\">5. Consider the velocity vs. time graph of a person in an elevator shown in Figure 13 Suppose the elevator is initially at rest. It then accelerates for 3 seconds, maintains that velocity for 15 seconds, then decelerates for 5 seconds until it stops. The acceleration for the entire trip is not constant so we cannot use the equations of motion from Motion Equations for Constant Acceleration in One Dimension for the complete trip. (We could, however, use them in the three individual sections where acceleration is a constant.) Sketch graphs of:\u00a0(a) position vs. time and\u00a0(b) acceleration vs. time for this trip.<\/div>\r\n<div class=\"title\"><\/div>\r\n<div class=\"body\">\r\n<div class=\"problem\">\r\n<div id=\"m42103-import-auto-id2006890\" class=\"figure\" title=\"Figure 2.51.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n<div class=\"wp-caption alignnone\">\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\/20101804\/Figure_03_08Sol_07.jpg\" alt=\"Line graph of velocity versus time. Line begins at the origin and has a positive slope until it reaches 3 meters per second at 3 seconds. The slope is then zero until 18 seconds, where it becomes negative until the line reaches a velocity of 0 at 23 seconds.\" width=\"350\" height=\"406\" \/> Figure 13.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n6. A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, velocity, and acceleration of the cylinder vs. time as it goes up and then down the plane.\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Problems &amp; Exercises<\/h3>\r\n<div id=\"m42103-fs-id1571006\" class=\"exercise labeled check-understanding\">\r\n\r\nNote: There is always uncertainty in numbers taken from graphs. If your answers differ from expected values, examine them to see if they are within data extraction uncertainties estimated by you.\r\n\r\n<\/div>\r\n1. a) By taking the slope of the curve in Figure 14, verify that the velocity of the jet car is 115 m\/s at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span>=20 s<\/span>. (b) By taking the slope of the curve at any point in Figure 15 verify that the jet car\u2019s acceleration is <span class=\"token\">5.0 m\/s<sup>2<\/sup><\/span>.\r\n<div id=\"m42103-fs-id4088406\" class=\"exercise problems-exercises\">\r\n<div class=\"body\">\r\n<div class=\"problem\">\r\n<div id=\"m42103-import-auto-id1798398\" class=\"figure\" title=\"Figure 2.51.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n<div class=\"wp-caption alignnone\">\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\/20101824\/Figure_02_08Sol_11.jpg\" alt=\"Line graph of position over time. Line has positive slope that increases over time.\" width=\"350\" height=\"395\" \/> Figure 14.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\"><\/div>\r\n<\/div>\r\n<div id=\"m42103-import-auto-id4101417\" class=\"figure\" title=\"Figure 2.51.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n<div class=\"wp-caption aligncenter\">\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\/20101826\/Figure_02_08Sol_12.jpg\" alt=\"Line graph of velocity versus time. Line is straight with a positive slope.\" width=\"350\" height=\"426\" \/> Figure 15.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"m42103-fs-id4012994\" class=\"cnx-gentext-exercise cnx-gentext-autogenerated\">\r\n\r\n2. Using approximate values, calculate the slope of the curve in Figure 16 to verify that the velocity at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span>=10.0 s<\/span> is 0.208 m\/s. Assume all values are known to 3 significant figures.\r\n<div class=\"body\">\r\n<div class=\"problem\">\r\n<div id=\"m42103-import-auto-id4122996\" class=\"figure\" title=\"Figure 2.51.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n<div class=\"wp-caption alignnone\">\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\/20101827\/Figure_02_08Sol_13.jpg\" alt=\"Line graph of position versus time. Line is straight with a positive slope.\" width=\"350\" height=\"362\" \/> Figure 16.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\">\r\n<div id=\"m42103-fs-id1372323\" class=\"exercise problems-exercises\">\r\n\r\n3. Using approximate values, calculate the slope of the curve in Figure 16 to verify that the velocity at t = 30.0 s is 0.238 m\/s. Assume all values are known to 3 significant figures.\r\n\r\n4.\u00a0By taking the slope of the curve in Figure 17, verify that the acceleration is 3.2 m\/s<sup>2<\/sup> at <em>t<\/em> = 10 s.<span data-type=\"media\" data-alt=\"Line graph of velocity versus time. Line has a positive slope that decreases over time until the line flattens out.\">\r\n<\/span>\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\/1322\/2015\/12\/03205610\/Figure_02_08Sol_14.jpg\" alt=\"Line graph of velocity versus time. Line has a positive slope that decreases over time until the line flattens out.\" width=\"350\" height=\"430\" data-media-type=\"image\/jpg\" \/> Figure 17.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n5. Construct the displacement graph for the subway shuttle train as shown in <a href=\".\/chapter\/2-4-acceleration\" target=\"_blank\">Figure 2 from Acceleration<\/a>\u00a0(shown again below).\u00a0Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"300\"]<img src=\"http:\/\/cnx.org\/resources\/955096ca5e1aa0b56fe9b6bfc7f67ae59e354380\/Figure_02_04_00a.jpg\" alt=\"A subway train arriving at a station. A velocity vector arrow points along the track away from the train. An acceleration vector arrow points along the track toward the train.\" width=\"300\" height=\"663\" data-media-type=\"image\/jpg\" \/> A subway train in Sao Paulo, Brazil, decelerates as it comes into a station. It is accelerating in a direction opposite to its direction of motion. (credit: Yusuke Kawasaki, Flickr)[\/caption]\r\n\r\n6. (a) Take the slope of the curve in Figure 11 to find the jogger\u2019s velocity at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>= 2.5 s<\/span>.\u00a0(b) Repeat at 7.5 s. These values must be consistent with the graph in Figure 18.\r\n\r\n<\/div>\r\n<div class=\"problem\">\r\n<div id=\"m42103-import-auto-id4122996\" class=\"figure\" title=\"Figure 2.51.\">\r\n<div class=\"title\">\r\n<div id=\"m42103-fs-id1372323\" class=\"exercise problems-exercises\"><\/div>\r\n<div id=\"m42103-fs-id2290187\" class=\"exercise problems-exercises\">\r\n<div class=\"body\">\r\n<div class=\"problem\">\r\n<div id=\"m42103-import-auto-id4064858\" class=\"figure\" title=\"Figure 2.51.\">\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\/20101833\/Figure_02_08Sol_16.jpg\" alt=\"Line graph of position over time. Line begins sloping upward, then kinks back down, then kinks back upward again.\" width=\"300\" height=\"389\" \/> Figure 18.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"m42103-import-auto-id4128350\" class=\"figure\" title=\"Figure 2.51.\">\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\/20101835\/Figure_02_08Sol_17.jpg\" alt=\"Line graph of velocity over time. Line begins with a positive slope, then kinks downward with a negative slope, then kinks back upward again. It kinks back down again slightly, then back up again, and ends with a slightly less positive slope.\" width=\"300\" height=\"461\" \/> Figure 19.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"title\"><\/div>\r\n<\/div>\r\n<div id=\"m42103-import-auto-id4151339\" class=\"figure\" title=\"Figure 2.51.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n<div class=\"wp-caption alignnone\">\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\/20101838\/Figure_02_08Sol_18.jpg\" alt=\"Graph depicting acceleration vs. time\" width=\"300\" height=\"407\" \/> Figure 20.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"m42103-fs-id3520768\" class=\"exercise problems-exercises\">\r\n<div class=\"body\">\r\n<div class=\"problem\">\r\n\r\n7. A graph of <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span>(<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span>)<\/span> is shown for a world-class track sprinter in a 100-m race. (See Figure 21).\u00a0(a) What is his average velocity for the first 4 s?\u00a0(b) What is his instantaneous velocity at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>= 5 s<\/span>?\u00a0(c) What is his average acceleration between 0 and 4 s?\u00a0(d) What is his time for the race?\r\n\r\n&nbsp;\r\n<div id=\"m42103-import-auto-id4125036\" class=\"figure\" title=\"Figure 2.51.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n<div class=\"wp-caption alignnone\">\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\/20101839\/Figure_02_08Sol_20.jpg\" alt=\"Line graph of velocity versus time. The line has two legs. The first has a constant positive slope. The second is flat, with a slope of 0.\" width=\"350\" height=\"444\" \/> Figure 21.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"m42103-fs-id1582774\" class=\"exercise problems-exercises\">\r\n<div class=\"body\">\r\n<div class=\"problem\">\r\n\r\n8. Figure 22 shows the displacement graph for a particle for 5 s. Draw the corresponding velocity and acceleration graphs.\r\n<div id=\"m42103-import-auto-id4035681\" class=\"figure\" title=\"Figure 2.51.\">\r\n<div class=\"body\">\r\n<div class=\"mediaobject\">\r\n<div class=\"wp-caption aligncenter\">\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\/20101841\/Figure_02_08Sol_21.jpg\" alt=\"Line graph of position versus time. The line has 4 legs. The first leg has a positive slope. The second leg has a negative slope. The third has a slope of 0. The fourth has a positive slope.\" width=\"350\" height=\"422\" \/> Figure 22.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"m42103-fs-id1770908\" class=\"exercise problems-exercises\"><\/div>\r\n<\/div>\r\n<div id=\"m42103-fs-id1770908\" class=\"exercise problems-exercises\">\r\n<div class=\"body\">\r\n<div class=\"problem\">\r\n<h2>Glossary<\/h2>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"glossary\">\r\n<dl id=\"fs-id1218535\" class=\"definition\">\r\n \t<dt>independent variable:<\/dt>\r\n \t<dd id=\"fs-id1343248\">the variable that the dependent variable is measured with respect to; usually plotted along the <em>x<\/em>-axis<\/dd>\r\n<\/dl>\r\n<dl id=\"import-auto-id1845390\" class=\"definition\">\r\n \t<dt>dependent variable:<\/dt>\r\n \t<dd id=\"fs-id4015576\">the variable that is being measured; usually plotted along the <em>y<\/em>-axis<\/dd>\r\n<\/dl>\r\n<dl id=\"import-auto-id3600469\" class=\"definition\">\r\n \t<dt>slope:<\/dt>\r\n \t<dd id=\"fs-id1544972\">the difference in <em>y<\/em>-value (the rise) divided by the difference in <em>x<\/em>-value (the run) of two points on a straight line<\/dd>\r\n<\/dl>\r\n<dl id=\"import-auto-id4021637\" class=\"definition\">\r\n \t<dt>y-intercept:<\/dt>\r\n \t<dd id=\"fs-id3525350\">the <em>y<\/em>-value when <em>x<\/em> = 0, or when the graph crosses the <em>y<\/em>-axis<\/dd>\r\n<\/dl>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Selected Solutions to Problems &amp; Exercises<\/h3>\r\n1. (a) 115 m\/s (b) 5.0 m\/s\r\n\r\n3.\u00a0[latex]v=\\frac{\\left(\\text{11.7}-6.95\\right)\\times {\\text{10}}^{3}\\text{m}}{\\left(40\\text{.}\\text{0 - 20}.0\\right)\\text{s}}=\\text{238 m\/s}\\\\[\/latex]\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\/1322\/2015\/12\/03205611\/Figure_02_08Sol_15.jpg\" alt=\"Line graph of position versus time. Line begins with a slight positive slope. It then kinks to a much greater positive slope.\" width=\"350\" height=\"429\" data-media-type=\"image\/jpg\" \/> Figure 23.[\/caption]\r\n\r\n5.\r\n\r\n7. (a) 6 m\/s\u00a0(b) 12 m\/s\u00a0(c) 3 m\/s<sup>2\u00a0<\/sup>(d) 10 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\">Describe a straight-line graph in terms of its slope and <span class=\"foreignphrase\"><em class=\"foreignphrase\">y<\/em><\/span>-intercept.<\/li>\n<li class=\"listitem\">Determine average velocity or instantaneous velocity from a graph of position vs. time.<\/li>\n<li class=\"listitem\">Determine average or instantaneous acceleration from a graph of velocity vs. time.<\/li>\n<li class=\"listitem\">Derive a graph of velocity vs. time from a graph of position vs. time.<\/li>\n<li class=\"listitem\">Derive a graph of acceleration vs. time from a graph of velocity vs. time.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p>A graph, like a picture, is worth a thousand words. Graphs not only contain numerical information; they also reveal relationships between physical quantities. This section uses graphs of displacement, velocity, and acceleration versus time to illustrate one-dimensional kinematics.<\/p>\n<div class=\"section\" title=\"Slopes and General Relationships\">\n<div class=\"titlepage\">\n<div>\n<div>\n<h2 id=\"m42103-fs-id1396690\"><span class=\"cnx-gentext-section cnx-gentext-t\">Slopes and General Relationships<\/span><\/h2>\n<\/div>\n<\/div>\n<\/div>\n<p>First note that graphs in this text have perpendicular axes, one horizontal and the other vertical. When two physical quantities are plotted against one another in such a graph, the horizontal axis is usually considered to be an <em class=\"glossterm\"> independent variable<\/em><a id=\"id783882\" class=\"indexterm\"><\/a> and the vertical axis a <em class=\"glossterm\"> dependent variable<\/em><a id=\"id783896\" class=\"indexterm\"><\/a>. If we call the horizontal axis the <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span>-axis and the vertical axis the <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>y<\/em><\/span><\/span>-axis, as in Figure 1, a straight-line graph has the general form<\/p>\n<div id=\"m42103-import-auto-id4175150\" class=\"equation\" title=\"Equation 2.101.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]y=\\text{mx}+b\\\\[\/latex]<\/div>\n<\/div>\n<p>Here <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>m<\/em><\/span><\/span> is the <em class=\"glossterm\"> slope<\/em><a id=\"id784470\" class=\"indexterm\"><\/a>, defined to be the rise divided by the run (as seen in the figure) of the straight line. The letter <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>b<\/em><\/span><\/span> is used for the <em class=\"glossterm\"> <span class=\"emphasis\"><em>y<\/em><\/span>-intercept<\/em><a id=\"id784596\" class=\"indexterm\"><\/a>, which is the point at which the line crosses the vertical axis.<\/p>\n<div id=\"m42103-import-auto-id2359358\" class=\"figure\" title=\"Figure 2.46.\">\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\/20101731\/Figure_02_07_01.jpg\" alt=\"Graph of a straight-line sloping up at about 40 degrees.\" width=\"300\" height=\"308\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. A straight-line graph. The equation for a straight line is .<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"section\" title=\"Graph of Displacement vs. Time (a = 0, so v is constant)\">\n<div class=\"titlepage\">\n<div>\n<div>\n<h2 id=\"m42103-fs-id2201114\"><span class=\"cnx-gentext-section cnx-gentext-t\">Graph of Displacement vs. Time (<span class=\"emphasis\"><em>a<\/em><\/span> = 0, so <span class=\"emphasis\"><em>v<\/em><\/span> is constant)<\/span><\/h2>\n<\/div>\n<\/div>\n<\/div>\n<p>Time is usually an independent variable that other quantities, such as displacement, depend upon. A graph of displacement versus time would, thus, have <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> on the vertical axis and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> on the horizontal axis. Figure 2\u00a0is just such a straight-line graph. It shows a graph of displacement versus time for a jet-powered car on a very flat dry lake bed in Nevada.<\/p>\n<div id=\"m42103-import-auto-id2574769\" class=\"figure\" title=\"Figure 2.47.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 460px\" 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\/20101733\/Figure_02_07_02.jpg\" alt=\"Line graph of jet car displacement in meters versus time in seconds. The line is straight with a positive slope. The y intercept is four hundred meters. The total change in time is eight point zero seconds. The initial position is four hundred meters. The final position is two thousand meters.\" width=\"450\" height=\"440\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. Graph of displacement versus time for a jet-powered car on the Bonneville Salt Flats.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Using the relationship between dependent and independent variables, we see that the slope in the graph above is average velocity [latex]\\bar{v}\\\\[\/latex]\u00a0and the intercept is displacement at time zero\u2014that is, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><sub>0<\/sub><\/span>. Substituting these symbols into [latex]y=\\text{mx}+b\\\\[\/latex]\u00a0gives<\/p>\n<div id=\"m42103-import-auto-id4019047\" class=\"equation\" title=\"Equation 2.102.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]x=\\bar{v}t+{x}_{0}\\\\[\/latex]<\/div>\n<\/div>\n<p>or<\/p>\n<div id=\"m42103-import-auto-id2357165\" class=\"equation\" title=\"Equation 2.103.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]x={x}_{0}+\\bar{v}t\\\\[\/latex].<\/div>\n<\/div>\n<p>Thus a graph of displacement versus time gives a general relationship among displacement, velocity, and time, as well as giving detailed numerical information about a specific situation.<\/p>\n<div id=\"m42103-fs-id4125096\" class=\"note\">\n<div class=\"title textbox shaded\">\n<h3 class=\"title\"><strong><span class=\"cnx-gentext-tip-t\">The Slope of <span class=\"emphasis\"><em>x<\/em><\/span> vs. <span class=\"emphasis\"><em>t<\/em><\/span> <\/span><\/strong><\/h3>\n<div class=\"body\">\n<p>The slope of the graph of displacement <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> vs. time <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> is velocity <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span>.<\/p>\n<p style=\"text-align: center;\">[latex]\\text{slope}=\\frac{\\Delta x}{\\Delta t}=v\\\\[\/latex]<\/p>\n<p>Notice that this equation is the same as that derived algebraically from other motion equations in <a class=\"link\" title=\"2.5. Motion Equations for Constant Acceleration in One Dimension\" href=\".\/chapter\/2-5-motion-equations-for-constant-acceleration-in-one-dimension\/\" target=\"_blank\">Motion Equations for Constant Acceleration in One Dimension<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>From the figure we can see that the car has a displacement of 400 m at time 0.650 m at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> = 1.0 s, and so on. Its displacement at times other than those listed in the table can be read from the graph; furthermore, information about its velocity and acceleration can also be obtained from the graph.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1. Determining Average Velocity from a Graph of Displacement versus Time: Jet Car<\/h3>\n<div class=\"body\">\n<p>Find the average velocity of the car whose position is graphed in Figure 2.<\/p>\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\n<p>The slope of a graph of <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> vs. <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> is average velocity, since slope equals rise over run. In this case, rise = change in displacement and run = change in time, so that<\/p>\n<div id=\"m42103-import-auto-id1850584\" class=\"equation\" title=\"Equation 2.105.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]\\text{slope}=\\frac{\\Delta x}{\\Delta t}=\\bar{v}\\\\[\/latex].<\/div>\n<div class=\"mediaobject\"><\/div>\n<\/div>\n<p>Since the slope is constant here, any two points on the graph can be used to find the slope. (Generally speaking, it is most accurate to use two widely separated points on the straight line. This is because any error in reading data from the graph is proportionally smaller if the interval is larger.)<\/p>\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\n<p>1. Choose two points on the line. In this case, we choose the points labeled on the graph: (6.4 s, 2000 m) and (0.50 s, 525 m). (Note, however, that you could choose any two points.)<\/p>\n<p>2. Substitute the <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> and <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> values of the chosen points into the equation. Remember in calculating change <span class=\"token\">(\u0394)<\/span> we always use final value minus initial value.<\/p>\n<div id=\"m42103-import-auto-id4181834\" class=\"equation\" title=\"Equation 2.106.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]\\bar{v}=\\frac{\\Delta x}{\\Delta t}=\\frac{\\text{2000 m}-\\text{525 m}}{6\\text{.}\\text{4 s}-0\\text{.}\\text{50 s}}\\\\[\/latex],<\/div>\n<\/div>\n<p>yielding<\/p>\n<div id=\"m42103-import-auto-id2024433\" class=\"equation\" title=\"Equation 2.107.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]\\bar{v}=\\text{250 m\/s}[\/latex].<\/div>\n<div class=\"mediaobject\"><\/div>\n<\/div>\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\n<p>This is an impressively large land speed (900 km\/h, or about 560 mi\/h): much greater than the typical highway speed limit of 60 mi\/h (27 m\/s or 96 km\/h), but considerably shy of the record of 343 m\/s (1234 km\/h or 766 mi\/h) set in 1997.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"section\" title=\"Graphs of Motion when a is constant but a\u22600\">\n<div class=\"titlepage\">\n<div>\n<div>\n<h2 id=\"m42103-fs-id2570304\"><span class=\"cnx-gentext-section cnx-gentext-t\">Graphs of Motion when <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span> is constant but <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span>\u22600<\/span><\/span><\/h2>\n<\/div>\n<\/div>\n<\/div>\n<p>The graphs in Figure 3\u00a0below represent the motion of the jet-powered car as it accelerates toward its top speed, but only during the time when its acceleration is constant. Time starts at zero for this motion (as if measured with a stopwatch), and the displacement and velocity are initially 200 m and 15 m\/s, respectively.<\/p>\n<div id=\"m42103-import-auto-id3596921\" class=\"figure\" title=\"Figure 2.48.\">\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\/20101738\/Figure_02_07_03.jpg\" alt=\"Three line graphs. First is a line graph of displacement over time. Line has a positive slope that increases with time. Second line graph is of velocity over time. Line is straight with a positive slope. Third line graph is of acceleration over time. Line is straight and horizontal, indicating constant acceleration.\" width=\"300\" height=\"1397\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. Graphs of motion of a jet-powered car during the time span when its acceleration is constant. (a) The slope of an x vs. t graph is velocity. This is shown at two points, and the instantaneous velocities obtained are plotted in the next graph. Instantaneous velocity at any point is the slope of the tangent at that point. (b) The slope of the v vs. t graph is constant for this part of the motion, indicating constant acceleration. (c) Acceleration has the constant value of 5.0 m\/s2 over the time interval plotted.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"title\"><\/div>\n<\/div>\n<div id=\"m42103-import-auto-id3583460\" class=\"figure\" title=\"Figure 2.49.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 310px\" class=\"wp-caption alignright\"><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\/20101740\/Figure_02_07_03a.jpg\" alt=\"Figure (Figure_02_07_03a.jpg)\" width=\"300\" height=\"450\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. A U.S. Air Force jet car speeds down a track. (credit: Matt Trostle, Flickr)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"title\">The graph of displacement versus time in Figure\u00a03(a) is a curve rather than a straight line. The slope of the curve becomes steeper as time progresses, showing that the velocity is increasing over time. The slope at any point on a displacement-versus-time graph is the instantaneous velocity at that point. It is found by drawing a straight line tangent to the curve at the point of interest and taking the slope of this straight line. Tangent lines are shown for two points in Figure 3(a). If this is done at every point on the curve and the values are plotted against time, then the graph of velocity versus time shown in Figure 3(b) is obtained. Furthermore, the slope of the graph of velocity versus time is acceleration, which is shown in Figure 3(c).<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 2. Determining Instantaneous Velocity from the Slope at a Point: Jet Car<\/h3>\n<div class=\"section\" title=\"Graphs of Motion when a is constant but a\u22600\">\n<div class=\"body\">\n<p>Calculate the velocity of the jet car at a time of 25 s by finding the slope of the <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> vs. <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> graph in the graph below.<\/p>\n<div id=\"m42103-import-auto-id4141386\" class=\"figure\" title=\"Figure 2.50.\">\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\/20101741\/graphics41.jpg\" alt=\"A graph of displacement versus time for a jet car. The x axis for time runs from zero to thirty five seconds. The y axis for displacement runs from zero to three thousand meters. The curve depicting displacement is concave up. The slope of the curve increases over time. Slope equals velocity v. There are two points on the curve, labeled, P and Q. P is located at time equals ten seconds. Q is located and time equals twenty-five seconds. A line tangent to P at ten seconds is drawn and has a slope delta x sub P over delta t sub p. A line tangent to Q at twenty five seconds is drawn and has a slope equal to delta x sub q over delta t sub q. Select coordinates are given in a table and consist of the following: time zero seconds displacement two hundred meters; time five seconds displacement three hundred thirty eight meters; time ten seconds displacement six hundred meters; time fifteen seconds displacement nine hundred eighty eight meters. Time twenty seconds displacement one thousand five hundred meters; time twenty five seconds displacement two thousand one hundred thirty eight meters; time thirty seconds displacement two thousand nine hundred meters.\" width=\"300\" height=\"368\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"title\"><\/div>\n<div class=\"caption\">The slope of an <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> vs. <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> graph is velocity. This is shown at two points. Instantaneous velocity at any point is the slope of the tangent at that point.<\/div>\n<div class=\"caption\"><\/div>\n<\/div>\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\n<p>The slope of a curve at a point is equal to the slope of a straight line tangent to the curve at that point. This principle is illustrated in Figure 5, where Q is the point at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span>=25 s<\/span>.<\/p>\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\n<p>1. Find the tangent line to the curve at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>= 25 s<\/span>.<\/p>\n<p>2. Determine the endpoints of the tangent. These correspond to a position of 1300 m at time 19 s and a position of 3120 m at time 32 s.<\/p>\n<p>3. Plug these endpoints into the equation to solve for the slope, <span class=\"emphasis\"><em><span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span><\/em><\/span>.<\/p>\n<div id=\"m42103-import-auto-id3627758\" class=\"equation\" title=\"Equation 2.108.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]\\text{slope}={v}_{Q}=\\frac{{\\Delta x}_{Q}}{{\\Delta t}_{Q}}=\\frac{\\left(\\text{3120 m}-\\text{1300 m}\\right)}{\\left(\\text{32 s}-\\text{19 s}\\right)}\\\\[\/latex]<\/div>\n<div class=\"mediaobject\"><\/div>\n<\/div>\n<p>Thus,<\/p>\n<p style=\"text-align: center;\">[latex]{v}_{Q}=\\frac{\\text{1820 m}}{\\text{13 s}}=\\text{140 m\/s.}\\\\[\/latex]<\/p>\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\n<p>This is the value given in this figure\u2019s table for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span> at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>= 25 s<\/span>. The value of 140 m\/s for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>Q<\/sub><\/span> is plotted in Figure 5. The entire graph of <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span> vs. <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> can be obtained in this fashion.<\/p>\n<\/div>\n<\/div>\n<p>Carrying this one step further, we note that the slope of a velocity versus time graph is acceleration. Slope is rise divided by run; on a <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span> vs. <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> graph, rise = change in velocity <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span> and run = change in time <span class=\"token\">\u0394<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span>.<\/p>\n<\/div>\n<\/div>\n<div id=\"m42103-fs-id1405001\" class=\"note\">\n<div class=\"title textbox shaded\">\n<h3 class=\"title\"><strong><span class=\"cnx-gentext-tip-t\">The Slope of <span class=\"emphasis\"><em>v<\/em><\/span> vs. <span class=\"emphasis\"><em>t<\/em><\/span><\/span><\/strong><\/h3>\n<div class=\"body\">\n<p>The slope of a graph of velocity <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span> vs. time <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> is acceleration <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span><span class=\"emphasis\"><em>.<\/em><\/span><\/p>\n<div id=\"m42103-import-auto-id4096826\" class=\"equation\" title=\"Equation 2.110.\">\n<div class=\"mediaobject\" style=\"text-align: center;\">[latex]\\text{slope}=\\frac{\\Delta v}{\\Delta t}=a\\\\[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Since the velocity versus time graph in Figure 3(b) is a straight line, its slope is the same everywhere, implying that acceleration is constant. Acceleration versus time is graphed in Figure 3(c).<\/p>\n<p>Additional general information can be obtained from Figure 5\u00a0and the expression for a straight line,[latex]y=\\text{mx}+b\\\\[\/latex].<\/p>\n<p>In this case, the vertical axis <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>y<\/em><\/span><\/span> is <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>V<\/em><\/span><\/span>, the intercept <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>b<\/em><\/span><\/span> is <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><sub>0<\/sub><\/span>, the slope <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>m<\/em><\/span><\/span> is <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span>, and the horizontal axis <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> is <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span>. Substituting these symbols yields<\/p>\n<p style=\"text-align: center;\">[latex]v={v}_{0}+\\text{at}\\\\[\/latex].<\/p>\n<p>A general relationship for velocity, acceleration, and time has again been obtained from a graph. Notice that this equation was also derived algebraically from other motion equations in <a class=\"link\" title=\"2.5. Motion Equations for Constant Acceleration in One Dimension\" href=\".\/chapter\/2-5-motion-equations-for-constant-acceleration-in-one-dimension\/\" target=\"_blank\">Motion Equations for Constant Acceleration in One Dimension<\/a>.<\/p>\n<p>It is not accidental that the same equations are obtained by graphical analysis as by algebraic techniques. In fact, an important way to <span class=\"emphasis\"><em>discover<\/em><\/span> physical relationships is to measure various physical quantities and then make graphs of one quantity against another to see if they are correlated in any way. Correlations imply physical relationships and might be shown by smooth graphs such as those above. From such graphs, mathematical relationships can sometimes be postulated. Further experiments are then performed to determine the validity of the hypothesized relationships.<\/p>\n<div class=\"section\" title=\"Graphs of Motion Where Acceleration is Not Constant\">\n<div class=\"titlepage\">\n<div>\n<div>\n<h2 id=\"m42103-fs-id2306208\"><span class=\"cnx-gentext-section cnx-gentext-t\">Graphs of Motion Where Acceleration is Not Constant<\/span><\/h2>\n<\/div>\n<\/div>\n<\/div>\n<p>Now consider the motion of the jet car as it goes from 165 m\/s to its top velocity of 250 m\/s, graphed in Figure 6. Time again starts at zero, and the initial displacement and velocity are 2900 m and 165 m\/s, respectively. (These were the final displacement and velocity of the car in the motion graphed in Figure 3.) Acceleration gradually decreases from <span class=\"token\">5.0 m\/s<sup>2<\/sup><\/span> to zero when the car hits 250 m\/s. The slope of the <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>x<\/em><\/span><\/span> vs. <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> graph increases until <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span>=55 s<\/span>, after which time the slope is constant. Similarly, velocity increases until 55 s and then becomes constant, since acceleration decreases to zero at 55 s and remains zero afterward.<\/p>\n<div id=\"m42103-import-auto-id1534076\" class=\"figure\" title=\"Figure 2.51.\">\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\/20101745\/Figure_02_07_04.jpg\" alt=\"Three line graphs of jet car displacement, velocity, and acceleration, respectively. First line graph is of position over time. Line is straight with a positive slope. Second line graph is of velocity over time. Line graph has a positive slope that decreases over time and flattens out at the end. Third line graph is of acceleration over time. Line has a negative slope that increases over time until it flattens out at the end. The line is not smooth, but has several kinks.\" width=\"350\" height=\"1327\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. Graphs of motion of a jet-powered car as it reaches its top velocity. This motion begins where the motion in Figure 3 ends. (a) The slope of this graph is velocity; it is plotted in the next graph. (b) The velocity gradually approaches its top value. The slope of this graph is acceleration; it is plotted in the final graph. (c) Acceleration gradually declines to zero when velocity becomes constant.<\/p>\n<\/div>\n<\/div>\n<\/div>\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\">\u00a0<\/span><span class=\"cnx-gentext-example cnx-gentext-t\">Calculating Acceleration from a Graph of Velocity versus Time<\/span><\/h3>\n<div id=\"m42103-import-auto-id1534076\" class=\"figure\" title=\"Figure 2.51.\">\n<div class=\"body\">\n<p>Calculate the acceleration of the jet car at a time of 25 s by finding the slope of the <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span><\/span> vs. <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><\/span> graph in Figure 6(b).<\/p>\n<h4><span class=\"bold\"><strong>Strategy<\/strong><\/span><\/h4>\n<p>The slope of the curve at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>= 25 s<\/span> is equal to the slope of the line tangent at that point, as illustrated in Figure 6(b).<\/p>\n<h4><span class=\"bold\"><strong>Solution<\/strong><\/span><\/h4>\n<p>Determine endpoints of the tangent line from the figure, and then plug them into the equation to solve for slope, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span>.<\/p>\n<div id=\"import-auto-id3503054\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\text{slope}=\\frac{\\Delta v}{\\Delta t}=\\frac{\\left(\\text{260 m\/s}-\\text{210 m\/s}\\right)}{\\left(\\text{51 s}-1.0 s\\right)}\\\\[\/latex]<\/div>\n<div id=\"import-auto-id2028886\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]a=\\frac{\\text{50 m\/s}}{\\text{50 s}}=1\\text{.}0 m{\\text{\/s}}^{2}\\\\[\/latex].<\/div>\n<div id=\"m42103-import-auto-id2028886\" class=\"equation\" title=\"Equation 2.113.\"><\/div>\n<h4><span class=\"bold\"><strong>Discussion<\/strong><\/span><\/h4>\n<p>Note that this value for <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>a<\/em><\/span><\/span> is consistent with the value plotted in Figure 6(c) at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>= 25 s<\/span>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>A graph of displacement versus time can be used to generate a graph of velocity versus time, and a graph of velocity versus time can be used to generate a graph of acceleration versus time. We do this by finding the slope of the graphs at every point. If the graph is linear (i.e., a line with a constant slope), it is easy to find the slope at any point and you have the slope for every point. Graphical analysis of motion can be used to describe both specific and general characteristics of kinematics. Graphs can also be used for other topics in physics. An important aspect of exploring physical relationships is to graph them and look for underlying relationships.<\/p>\n<div id=\"m42103-fs-id1571006\" class=\"exercise labeled check-understanding\">\n<div class=\"textbox key-takeaways\">\n<div id=\"m42103-fs-id1571006\" class=\"exercise labeled check-understanding\">\n<h3 class=\"title\">Check Your Understanding<\/h3>\n<p>A graph of velocity vs. time of a ship coming into a harbor is shown below. (a) Describe the motion of the ship based on the graph. (b)What would a graph of the ship\u2019s acceleration look like?<\/p>\n<div id=\"m42103-import-auto-id3504346\" class=\"figure\" title=\"Figure 2.51.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 210px\" 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\/20101750\/Figure_02_07_04a.jpg\" alt=\"Line graph of velocity versus time. The line has three legs. The first leg is flat. The second leg has a negative slope. The third leg also has a negative slope, but the slope is not as negative as the second leg.\" width=\"200\" height=\"193\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"m42103-fs-id1658952\" class=\"solution\">\n<h4 class=\"title\"><strong>Solution<\/strong><\/h4>\n<div class=\"title\"><\/div>\n<div class=\"body\">\n<p>(a) The ship moves at constant velocity and then begins to decelerate at a constant rate. At some point, its deceleration rate decreases. It maintains this lower deceleration rate until it stops moving.<\/p>\n<p>(b) A graph of acceleration vs. time would show zero acceleration in the first leg, large and constant negative acceleration in the second leg, and constant negative acceleration.<\/p>\n<div id=\"m42103-import-auto-id1666671\" class=\"figure\" title=\"Figure 2.51.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div style=\"width: 210px\" 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\/20101752\/Figure_02_07_04b.jpg\" alt=\"A line graph of acceleration versus time. There are three legs of the graph. All three legs are flat and straight. The first leg shows constant acceleration of 0. The second leg shows a constant negative acceleration. The third leg shows a constant negative acceleration that is not as negative as the second leg.\" width=\"200\" height=\"193\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 8.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"title\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>Section Summary<\/h2>\n<ul>\n<li>Graphs of motion can be used to analyze motion.<\/li>\n<li>Graphical solutions yield identical solutions to mathematical methods for deriving motion equations.<\/li>\n<li>The slope of a graph of displacement <em>x<\/em> vs. time\u00a0<em>t<\/em>\u00a0is velocity <em>v<\/em>.<\/li>\n<li>The slope of a graph of velocity <em>v<\/em> vs. <em>t<\/em> graph is acceleration<em> a<\/em>.<\/li>\n<li>Average velocity, instantaneous velocity, and acceleration can all be obtained by analyzing graphs.<\/li>\n<\/ul>\n<div class=\"textbox key-takeaways\">\n<h3>Conceptual Questions<\/h3>\n<div id=\"m42103-fs-id1571006\" class=\"exercise labeled check-understanding\">\n<p>1. (a) Explain how you can use the graph of position versus time in Figure 9 to describe the change in velocity over time. Identify:\u00a0(b) the time (<span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>a<\/sub><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>b<\/sub><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>c<\/sub><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>d<\/sub><\/span>, or <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>e<\/sub><\/span>) at which the instantaneous velocity is greatest,\u00a0(c) the time at which it is zero, and\u00a0(d) the time at which it is negative.<\/p>\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\/1322\/2015\/12\/03205558\/Figure_03_08Sol_01.jpg\" alt=\"Line graph of position versus time with 5 points labeled: a, b, c, d, and e. The slope of the line changes. It begins with a positive slope that decreases over time until around point d, where it is flat. It then has a slightly negative slope.\" width=\"300\" height=\"411\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 9.<\/p>\n<\/div>\n<p>2. (a) Sketch a graph of velocity versus time corresponding to the graph of displacement versus time given in Figure 10.\u00a0(b) Identify the time or times (<span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>a<\/sub><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>b<\/sub><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>c<\/sub><\/span>, etc.) at which the instantaneous velocity is greatest.\u00a0(c) At which times is it zero?\u00a0(d) At which times is it negative?<\/p>\n<\/div>\n<div class=\"title\"><\/div>\n<div class=\"body\">\n<div class=\"problem\">\n<div id=\"m42103-import-auto-id2562897\" class=\"figure\" title=\"Figure 2.51.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div class=\"wp-caption alignnone\">\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\/20101800\/Figure_03_08Sol_02.jpg\" alt=\"Line graph of position over time with 12 points labeled a through l. Line has a negative slope from a to c, where it turns and has a positive slope till point e. It turns again and has a negative slope till point g. The slope then increases again till l, where it flattens out.\" width=\"300\" height=\"416\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 10.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>3. (a) Explain how you can determine the acceleration over time from a velocity versus time graph such as the one in Figure 11.\u00a0(b) Based on the graph, how does acceleration change over time?<\/p>\n<div class=\"body\">\n<div class=\"problem\">\n<div id=\"m42103-import-auto-id1778975\" class=\"figure\" title=\"Figure 2.51.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div class=\"wp-caption alignnone\">\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\/20101801\/Figure_03_08Sol_04.jpg\" alt=\"Line graph of velocity over time with two points labeled. Point P is at v 1 t 1. Point Q is at v 2 t 2. The line has a positive slope that increases over time.\" width=\"300\" height=\"391\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 11.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>4. \u00a0(a) Sketch a graph of acceleration versus time corresponding to the graph of velocity versus time given in Figure 12.\u00a0(b) Identify the time or times (<span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>a<\/sub><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>b<\/sub><\/span>, <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span><sub>c<\/sub><\/span>, etc.) at which the acceleration is greatest.\u00a0(c) At which times is it zero?\u00a0(d) At which times is it negative?<\/p>\n<div class=\"body\">\n<div class=\"problem\">\n<div id=\"m42103-import-auto-id1447833\" class=\"figure\" title=\"Figure 2.51.\">\n<div class=\"body\">\n<div class=\"wp-caption alignnone\">\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\/20101802\/Figure_03_08Sol_05.jpg\" alt=\"Line graph of velocity over time with 12 points labeled a through l. The line has a positive slope from a at the origin to d where it slopes downward to e, and then back upward to h. It then slopes back down to point l at v equals 0.\" width=\"300\" height=\"408\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 12.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"title\">5. Consider the velocity vs. time graph of a person in an elevator shown in Figure 13 Suppose the elevator is initially at rest. It then accelerates for 3 seconds, maintains that velocity for 15 seconds, then decelerates for 5 seconds until it stops. The acceleration for the entire trip is not constant so we cannot use the equations of motion from Motion Equations for Constant Acceleration in One Dimension for the complete trip. (We could, however, use them in the three individual sections where acceleration is a constant.) Sketch graphs of:\u00a0(a) position vs. time and\u00a0(b) acceleration vs. time for this trip.<\/div>\n<div class=\"title\"><\/div>\n<div class=\"body\">\n<div class=\"problem\">\n<div id=\"m42103-import-auto-id2006890\" class=\"figure\" title=\"Figure 2.51.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div class=\"wp-caption alignnone\">\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\/20101804\/Figure_03_08Sol_07.jpg\" alt=\"Line graph of velocity versus time. Line begins at the origin and has a positive slope until it reaches 3 meters per second at 3 seconds. The slope is then zero until 18 seconds, where it becomes negative until the line reaches a velocity of 0 at 23 seconds.\" width=\"350\" height=\"406\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 13.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>6. A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, velocity, and acceleration of the cylinder vs. time as it goes up and then down the plane.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Problems &amp; Exercises<\/h3>\n<div id=\"m42103-fs-id1571006\" class=\"exercise labeled check-understanding\">\n<p>Note: There is always uncertainty in numbers taken from graphs. If your answers differ from expected values, examine them to see if they are within data extraction uncertainties estimated by you.<\/p>\n<\/div>\n<p>1. a) By taking the slope of the curve in Figure 14, verify that the velocity of the jet car is 115 m\/s at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span>=20 s<\/span>. (b) By taking the slope of the curve at any point in Figure 15 verify that the jet car\u2019s acceleration is <span class=\"token\">5.0 m\/s<sup>2<\/sup><\/span>.<\/p>\n<div id=\"m42103-fs-id4088406\" class=\"exercise problems-exercises\">\n<div class=\"body\">\n<div class=\"problem\">\n<div id=\"m42103-import-auto-id1798398\" class=\"figure\" title=\"Figure 2.51.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div class=\"wp-caption alignnone\">\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\/20101824\/Figure_02_08Sol_11.jpg\" alt=\"Line graph of position over time. Line has positive slope that increases over time.\" width=\"350\" height=\"395\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 14.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"title\"><\/div>\n<\/div>\n<div id=\"m42103-import-auto-id4101417\" class=\"figure\" title=\"Figure 2.51.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div class=\"wp-caption aligncenter\">\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\/20101826\/Figure_02_08Sol_12.jpg\" alt=\"Line graph of velocity versus time. Line is straight with a positive slope.\" width=\"350\" height=\"426\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 15.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"m42103-fs-id4012994\" class=\"cnx-gentext-exercise cnx-gentext-autogenerated\">\n<p>2. Using approximate values, calculate the slope of the curve in Figure 16 to verify that the velocity at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t<\/em><\/span>=10.0 s<\/span> is 0.208 m\/s. Assume all values are known to 3 significant figures.<\/p>\n<div class=\"body\">\n<div class=\"problem\">\n<div id=\"m42103-import-auto-id4122996\" class=\"figure\" title=\"Figure 2.51.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div class=\"wp-caption alignnone\">\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\/20101827\/Figure_02_08Sol_13.jpg\" alt=\"Line graph of position versus time. Line is straight with a positive slope.\" width=\"350\" height=\"362\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 16.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"title\">\n<div id=\"m42103-fs-id1372323\" class=\"exercise problems-exercises\">\n<p>3. Using approximate values, calculate the slope of the curve in Figure 16 to verify that the velocity at t = 30.0 s is 0.238 m\/s. Assume all values are known to 3 significant figures.<\/p>\n<p>4.\u00a0By taking the slope of the curve in Figure 17, verify that the acceleration is 3.2 m\/s<sup>2<\/sup> at <em>t<\/em> = 10 s.<span data-type=\"media\" data-alt=\"Line graph of velocity versus time. Line has a positive slope that decreases over time until the line flattens out.\"><br \/>\n<\/span><\/p>\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\/1322\/2015\/12\/03205610\/Figure_02_08Sol_14.jpg\" alt=\"Line graph of velocity versus time. Line has a positive slope that decreases over time until the line flattens out.\" width=\"350\" height=\"430\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 17.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>5. Construct the displacement graph for the subway shuttle train as shown in <a href=\".\/chapter\/2-4-acceleration\" target=\"_blank\">Figure 2 from Acceleration<\/a>\u00a0(shown again below).\u00a0Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.<\/p>\n<div style=\"width: 310px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/cnx.org\/resources\/955096ca5e1aa0b56fe9b6bfc7f67ae59e354380\/Figure_02_04_00a.jpg\" alt=\"A subway train arriving at a station. A velocity vector arrow points along the track away from the train. An acceleration vector arrow points along the track toward the train.\" width=\"300\" height=\"663\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\">A subway train in Sao Paulo, Brazil, decelerates as it comes into a station. It is accelerating in a direction opposite to its direction of motion. (credit: Yusuke Kawasaki, Flickr)<\/p>\n<\/div>\n<p>6. (a) Take the slope of the curve in Figure 11 to find the jogger\u2019s velocity at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>= 2.5 s<\/span>.\u00a0(b) Repeat at 7.5 s. These values must be consistent with the graph in Figure 18.<\/p>\n<\/div>\n<div class=\"problem\">\n<div id=\"m42103-import-auto-id4122996\" class=\"figure\" title=\"Figure 2.51.\">\n<div class=\"title\">\n<div id=\"m42103-fs-id1372323\" class=\"exercise problems-exercises\"><\/div>\n<div id=\"m42103-fs-id2290187\" class=\"exercise problems-exercises\">\n<div class=\"body\">\n<div class=\"problem\">\n<div id=\"m42103-import-auto-id4064858\" class=\"figure\" title=\"Figure 2.51.\">\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\/20101833\/Figure_02_08Sol_16.jpg\" alt=\"Line graph of position over time. Line begins sloping upward, then kinks back down, then kinks back upward again.\" width=\"300\" height=\"389\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 18.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"m42103-import-auto-id4128350\" class=\"figure\" title=\"Figure 2.51.\">\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\/20101835\/Figure_02_08Sol_17.jpg\" alt=\"Line graph of velocity over time. Line begins with a positive slope, then kinks downward with a negative slope, then kinks back upward again. It kinks back down again slightly, then back up again, and ends with a slightly less positive slope.\" width=\"300\" height=\"461\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 19.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"title\"><\/div>\n<\/div>\n<div id=\"m42103-import-auto-id4151339\" class=\"figure\" title=\"Figure 2.51.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div class=\"wp-caption alignnone\">\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\/20101838\/Figure_02_08Sol_18.jpg\" alt=\"Graph depicting acceleration vs. time\" width=\"300\" height=\"407\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 20.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"m42103-fs-id3520768\" class=\"exercise problems-exercises\">\n<div class=\"body\">\n<div class=\"problem\">\n<p>7. A graph of <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>v<\/em><\/span>(<span class=\"emphasis mathml-mi\"><em>t<\/em><\/span>)<\/span> is shown for a world-class track sprinter in a 100-m race. (See Figure 21).\u00a0(a) What is his average velocity for the first 4 s?\u00a0(b) What is his instantaneous velocity at <span class=\"token\"><span class=\"emphasis mathml-mi\"><em>t\u00a0<\/em><\/span>= 5 s<\/span>?\u00a0(c) What is his average acceleration between 0 and 4 s?\u00a0(d) What is his time for the race?<\/p>\n<p>&nbsp;<\/p>\n<div id=\"m42103-import-auto-id4125036\" class=\"figure\" title=\"Figure 2.51.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div class=\"wp-caption alignnone\">\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\/20101839\/Figure_02_08Sol_20.jpg\" alt=\"Line graph of velocity versus time. The line has two legs. The first has a constant positive slope. The second is flat, with a slope of 0.\" width=\"350\" height=\"444\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 21.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"m42103-fs-id1582774\" class=\"exercise problems-exercises\">\n<div class=\"body\">\n<div class=\"problem\">\n<p>8. Figure 22 shows the displacement graph for a particle for 5 s. Draw the corresponding velocity and acceleration graphs.<\/p>\n<div id=\"m42103-import-auto-id4035681\" class=\"figure\" title=\"Figure 2.51.\">\n<div class=\"body\">\n<div class=\"mediaobject\">\n<div class=\"wp-caption aligncenter\">\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\/20101841\/Figure_02_08Sol_21.jpg\" alt=\"Line graph of position versus time. The line has 4 legs. The first leg has a positive slope. The second leg has a negative slope. The third has a slope of 0. The fourth has a positive slope.\" width=\"350\" height=\"422\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 22.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"m42103-fs-id1770908\" class=\"exercise problems-exercises\"><\/div>\n<\/div>\n<div id=\"m42103-fs-id1770908\" class=\"exercise problems-exercises\">\n<div class=\"body\">\n<div class=\"problem\">\n<h2>Glossary<\/h2>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"glossary\">\n<dl id=\"fs-id1218535\" class=\"definition\">\n<dt>independent variable:<\/dt>\n<dd id=\"fs-id1343248\">the variable that the dependent variable is measured with respect to; usually plotted along the <em>x<\/em>-axis<\/dd>\n<\/dl>\n<dl id=\"import-auto-id1845390\" class=\"definition\">\n<dt>dependent variable:<\/dt>\n<dd id=\"fs-id4015576\">the variable that is being measured; usually plotted along the <em>y<\/em>-axis<\/dd>\n<\/dl>\n<dl id=\"import-auto-id3600469\" class=\"definition\">\n<dt>slope:<\/dt>\n<dd id=\"fs-id1544972\">the difference in <em>y<\/em>-value (the rise) divided by the difference in <em>x<\/em>-value (the run) of two points on a straight line<\/dd>\n<\/dl>\n<dl id=\"import-auto-id4021637\" class=\"definition\">\n<dt>y-intercept:<\/dt>\n<dd id=\"fs-id3525350\">the <em>y<\/em>-value when <em>x<\/em> = 0, or when the graph crosses the <em>y<\/em>-axis<\/dd>\n<\/dl>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Selected Solutions to Problems &amp; Exercises<\/h3>\n<p>1. (a) 115 m\/s (b) 5.0 m\/s<\/p>\n<p>3.\u00a0[latex]v=\\frac{\\left(\\text{11.7}-6.95\\right)\\times {\\text{10}}^{3}\\text{m}}{\\left(40\\text{.}\\text{0 - 20}.0\\right)\\text{s}}=\\text{238 m\/s}\\\\[\/latex]<\/p>\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\/1322\/2015\/12\/03205611\/Figure_02_08Sol_15.jpg\" alt=\"Line graph of position versus time. Line begins with a slight positive slope. It then kinks to a much greater positive slope.\" width=\"350\" height=\"429\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 23.<\/p>\n<\/div>\n<p>5.<\/p>\n<p>7. (a) 6 m\/s\u00a0(b) 12 m\/s\u00a0(c) 3 m\/s<sup>2\u00a0<\/sup>(d) 10 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-342\">\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\/Ax2o07Ul@9.4:7x6Qspwt@6\/Graphical-Analysis-of-One-Dime\">http:\/\/cnx.org\/contents\/Ax2o07Ul@9.4:7x6Qspwt@6\/Graphical-Analysis-of-One-Dime<\/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\/031da8d3-b525-429c-80cf-6c8ed997733a@9.4.<\/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":10,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"College Physics\",\"author\":\"OpenStax College\",\"organization\":\"\",\"url\":\"http:\/\/cnx.org\/contents\/Ax2o07Ul@9.4:7x6Qspwt@6\/Graphical-Analysis-of-One-Dime\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/031da8d3-b525-429c-80cf-6c8ed997733a@9.4.\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-342","chapter","type-chapter","status-publish","hentry"],"part":7456,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/pressbooks\/v2\/chapters\/342","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/wp\/v2\/users\/1"}],"version-history":[{"count":21,"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/pressbooks\/v2\/chapters\/342\/revisions"}],"predecessor-version":[{"id":12009,"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/pressbooks\/v2\/chapters\/342\/revisions\/12009"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/pressbooks\/v2\/parts\/7456"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/pressbooks\/v2\/chapters\/342\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/wp\/v2\/media?parent=342"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/pressbooks\/v2\/chapter-type?post=342"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/wp\/v2\/contributor?post=342"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/wp\/v2\/license?post=342"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}