{"id":359,"date":"2018-02-06T15:33:04","date_gmt":"2018-02-06T15:33:04","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/?post_type=chapter&#038;p=359"},"modified":"2018-07-04T14:53:01","modified_gmt":"2018-07-04T14:53:01","slug":"1-1-position-displacement-and-average-velocity","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/chapter\/1-1-position-displacement-and-average-velocity\/","title":{"raw":"3.1 Position, Displacement, and Average Velocity","rendered":"3.1 Position, Displacement, and Average Velocity"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Define position, displacement, and distance traveled.<\/li>\r\n \t<li>Calculate the total displacement given the position as a function of time.<\/li>\r\n \t<li>Determine the total distance traveled.<\/li>\r\n \t<li>Calculate the average velocity given the displacement and elapsed time.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1168329489850\">When you\u2019re in motion, the basic questions to ask are: Where are you? Where are you going? How fast are you getting there? The answers to these questions require that you specify your position, your displacement, and your average velocity\u2014the terms we define in this section.<\/p>\r\n\r\n<div id=\"fs-id1168329158454\" class=\"bc-section section\">\r\n<h3>Position<\/h3>\r\n<p id=\"fs-id1168326782871\">To describe the motion of an object, you must first be able to describe its<strong> position<\/strong> (<em>x<\/em>): <em>where it is at any particular time<\/em>. More precisely, we need to specify its position relative to a convenient frame of reference. A<strong> <span class=\"no-emphasis\">frame of reference<\/span><\/strong> is an arbitrary set of axes from which the position and motion of an object are described. Earth is often used as a frame of reference, and we often describe the position of an object as it relates to stationary objects on Earth. For example, a rocket launch could be described in terms of the position of the rocket with respect to Earth as a whole, whereas a cyclist\u2019s position could be described in terms of where she is in relation to the buildings she passes <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_03_01_Cyclists\">(Figure)<\/a>. In other cases, we use reference frames that are not stationary but are in motion relative to Earth. To describe the position of a person in an airplane, for example, we use the airplane, not Earth, as the reference frame. To describe the position of an object undergoing one-dimensional motion, we often use the variable <em>x<\/em>. Later in the chapter, during the discussion of free fall, we use the variable <em>y<\/em>.<\/p>\r\n\r\n<div id=\"CNX_UPhysics_03_01_Cyclists\" class=\"wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184959\/CNX_UPhysics_03_01_Cyclists.jpg\" alt=\"Picture shows three people riding bicycles next to a canal.\" width=\"487\" height=\"366\" \/> <strong>Figure 3.2<\/strong> These cyclists in Vietnam can be described by their position relative to buildings or a canal. Their motion can be described by their change in position, or displacement, in a frame of reference. (credit: Suzan Black)[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168326761623\" class=\"bc-section section\">\r\n<h3>Displacement<\/h3>\r\n<p id=\"fs-id1168329187699\">If an object moves relative to a frame of reference\u2014for example, if a professor moves to the right relative to a whiteboard <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_03_01_Chalkboard\">(Figure)<\/a>\u2014then the object\u2019s position changes. This change in position is called <strong>displacement<\/strong>. The word <em>displacement<\/em> implies that an object has moved, or has been displaced. Although position is the numerical value of <em>x<\/em> along a straight line where an object might be located, displacement gives the <em>change<\/em> in position along this line. Since displacement indicates direction, it is a vector and can be either positive or negative, depending on the choice of positive direction. Also, an analysis of motion can have many displacements embedded in it. If right is positive and an object moves 2 m to the right, then 4 m to the left, the individual displacements are 2 m and $$ -4 $$ m, respectively.<\/p>\r\n\r\n<div id=\"CNX_UPhysics_03_01_Chalkboard\" class=\"wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"689\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31185003\/CNX_UPhysics_03_01_Chalkboard.jpg\" alt=\"Illustration shows professor at two different locations. The first location is marked as 1.5 meters at the x axis; the second location is marked as 3.5 meters at the x axis. The displacement between the two locations is 2 meters.\" width=\"689\" height=\"544\" \/> <strong>Figure 3.3<\/strong> A professor paces left and right while lecturing. Her position relative to Earth is given by x. The +2.0-m displacement of the professor relative to Earth is represented by an arrow pointing to the right.[\/caption]\r\n\r\n<\/div>\r\n<div id=\"fs-id1168328983002\">\r\n<div><strong>Displacement<\/strong><\/div>\r\n<p id=\"fs-id1168327144862\">Displacement $$ \\text{\u0394}x $$ is the change in position of an object:<\/p>\r\n\r\n<div id=\"fs-id1168329338892\">$$\\text{\u0394}x={x}_{\\text{f}}-{x}_{0},$$<\/div>\r\n<p id=\"fs-id1168326911738\">where $$ \\text{\u0394}x $$ is displacement, $$ {x}_{\\text{f}} $$ is the final position, and $$ {x}_{0} $$ is the initial position.<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1168326795409\">We use the uppercase Greek letter delta (\u0394) to mean \u201cchange in\u201d whatever quantity follows it; thus, $$ \\text{\u0394}x $$ means <em>change in position<\/em> (final position less initial position). We always solve for displacement by subtracting initial position $$ {x}_{0} $$ from final position $$ {x}_{\\text{f}}$$. Note that the SI unit for displacement is the meter, but sometimes we use kilometers or other units of length. Keep in mind that when units other than meters are used in a problem, you may need to convert them to meters to complete the calculation (see <a href=\"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/back-matter\/conversion-factors\/\">Conversion Factors<\/a>).<\/p>\r\n<p id=\"fs-id1168329464556\">Objects in motion can also have a series of displacements. In the previous example of the pacing professor, the individual displacements are 2 m and $$ -4 $$ m, giving a total displacement of \u22122 m. We define<strong> total displacement<\/strong> $$ \\text{\u0394}{x}_{\\text{Total}}$$, as <em>the sum of the individual displacements<\/em>, and express this mathematically with the equation<\/p>\r\n\r\n<div class=\"equation-callout\">\r\n<div id=\"fs-id1168326821763\">$$\\text{\u0394}{x}_{\\text{Total}}=\\sum \\text{\u0394}{x}_{\\text{i}},$$<\/div>\r\n<\/div>\r\n<p id=\"fs-id1168329069172\">where $$ \\text{\u0394}{x}_{i} $$ are the individual displacements. In the earlier example,<\/p>\r\n\r\n<div id=\"fs-id1168326796770\" class=\"unnumbered\">$$\\text{\u0394}{x}_{1}={x}_{1}-{x}_{0}=2-0=2\\,\\text{m.}$$<\/div>\r\n<p id=\"fs-id1168329410562\">Similarly,<\/p>\r\n\r\n<div id=\"fs-id1168329155289\" class=\"unnumbered\">$$\\text{\u0394}{x}_{2}={x}_{2}-{x}_{1}=-2-(2)=-4\\,\\text{m.}$$<\/div>\r\n<p id=\"fs-id1168329081791\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1168326753395\" class=\"unnumbered\">$$\\text{\u0394}{x}_{\\text{Total}}=\\text{\u0394}{x}_{1}+\\text{\u0394}{x}_{2}=2-4=-2\\,\\text{m}\\text{\u200b.}$$<\/div>\r\n<p id=\"fs-id1168329518900\">The total displacement is 2 \u2212 4 = \u22122 m to the left, or in the negative direction. It is also useful to calculate the magnitude of the displacement, or its size. The magnitude of the displacement is always positive. This is the absolute value of the displacement, because displacement is a vector and cannot have a negative value of magnitude. In our example, the magnitude of the total displacement is 2 m, whereas the magnitudes of the individual displacements are 2 m and 4 m.<\/p>\r\n<p id=\"fs-id1168326908350\">The magnitude of the total displacement should not be confused with the distance traveled. Distance traveled $$ {x}_{\\text{Total}}$$, is the total length of the path traveled between two positions. In the previous problem, the <strong>distance traveled<\/strong> is the sum of the magnitudes of the individual displacements:<\/p>\r\n\r\n<div id=\"fs-id1168329266306\" class=\"unnumbered\">$${x}_{\\text{Total}}=|\\text{\u0394}{x}_{1}|+|\\text{\u0394}{x}_{2}|=2+4=6\\,\\text{m}\\text{.}$$<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168326783743\" class=\"bc-section section\">\r\n<h3>Average Velocity<\/h3>\r\n<p id=\"fs-id1168329046092\">To calculate the other physical quantities in kinematics we must introduce the time variable. The time variable allows us not only to state where the object is (its position) during its motion, but also how fast it is moving. How fast an object is moving is given by the rate at which the position changes with time.<\/p>\r\n<p id=\"fs-id1168329333293\">For each position $$ {x}_{\\text{i}}$$, we assign a particular time $$ {t}_{\\text{i}}$$. If the details of the motion at each instant are not important, the rate is usually expressed as the <strong>average velocity<\/strong> $$ \\overset{\\text{\u2013}}{v}$$. This vector quantity is simply the total displacement between two points divided by the time taken to travel between them. The time taken to travel between two points is called the <strong>elapsed time<\/strong> $$ \\text{\u0394}t$$.<\/p>\r\n\r\n<div id=\"fs-id1168329050548\">\r\n<div>Average Velocity<\/div>\r\n<p id=\"fs-id1168329192944\">If $$ {x}_{1} $$ and $$ {x}_{2} $$ are the positions of an object at times $$ {t}_{1} $$ and $$ {t}_{2}$$, respectively, then<\/p>\r\n\r\n<div id=\"fs-id1168326922062\">$$\\begin{array}{cc} \\text{Average velocity}=\\overset{\\text{\u2013}}{v}=\\frac{\\text{Displacement between two points}}{\\text{Elapsed time between two points}}\\\\ \\overset{\\text{\u2013}}{v}=\\frac{\\text{\u0394}x}{\\text{\u0394}t}=\\frac{{x}_{2}-{x}_{1}}{{t}_{2}-{t}_{1}}.\\end{array}$$<\/div>\r\n<\/div>\r\n<p id=\"fs-id1168326875739\">It is important to note that the average velocity is a vector and can be negative, depending on positions $$ {x}_{1} $$ and $$ {x}_{2}$$.<\/p>\r\n\r\n<div id=\"fs-id1168329011730\" class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\n<h4>Delivering Flyers<\/h4>\r\nJill sets out from her home to deliver flyers for her yard sale, traveling due east along her street lined with houses. At $$ 0.5 $$ km and 9 minutes later she runs out of flyers and has to retrace her steps back to her house to get more. This takes an additional 9 minutes. After picking up more flyers, she sets out again on the same path, continuing where she left off, and ends up 1.0 km from her house. This third leg of her trip takes $$ 15 $$ minutes. At this point she turns back toward her house, heading west. After $$ 1.75 $$ km and $$ 25 $$ minutes she stops to rest.\r\n<ol id=\"fs-id1168329101865\" type=\"a\">\r\n \t<li>What is Jill\u2019s total displacement to the point where she stops to rest?<\/li>\r\n \t<li>What is the magnitude of the final displacement?<\/li>\r\n \t<li>What is the average velocity during her entire trip?<\/li>\r\n \t<li>What is the total distance traveled?<\/li>\r\n \t<li>Make a graph of position versus time.<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1168329133729\">A sketch of Jill\u2019s movements is shown in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_03_01_Flyers\">(Figure)<\/a>.<\/p>\r\n\r\n<div id=\"CNX_UPhysics_03_01_Flyers\" class=\"wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"785\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31185006\/CNX_UPhysics_03_01_Flyers.jpg\" alt=\"Figure shows a timeline of a person\u2019s movement. First displacement is from the home to the right by 0.5 kilometers. Second displacement is back to the starting point. Third displacement is to the right by 1.0 kilometer. Fourth displacement is from the final point to the left by 1.75 kilometers.\" width=\"785\" height=\"333\" \/> Figure 3.4 Timeline of Jill\u2019s movements.[\/caption]\r\n\r\n<\/div>\r\n<h4>Strategy<\/h4>\r\nThe problem contains data on the various legs of Jill\u2019s trip, so it would be useful to make a table of the physical quantities. We are given position and time in the wording of the problem so we can calculate the displacements and the elapsed time. We take east to be the positive direction. From this information we can find the total displacement and average velocity. Jill\u2019s home is the starting point $$ {x}_{0}$$. The following table gives Jill\u2019s time and position in the first two columns, and the displacements are calculated in the third column.\r\n<table id=\"fs-id1168326821006\" class=\"unnumbered\" summary=\"This table has three columns and six rows. The first row is a header row and it labels each column: \u201cTime ti in minutes\u201d, \u201cPosition xi in kilometers\u201d, \u201cDisplacement Delta xi in kilometers\u201d. Under the \u201cTime ti in minutes\u201d: t0 = 0; t1 = 9; t2 = 18; t3 = 33; t4 = 58; Under the \u201cPosition xi in kilometers\u201d: x0 = 0; x1 = 0.5; x2 = 0; x3 = 1.0; x4 = -0.75; Under the \u201cDisplacement Delta xi in kilometers\u201d: Delta x0 = 0; Delta x1 = x1 \u2013 x0 = 0.5; Delta x2 = x2 \u2013 x1 = -0.5; Delta x3 = x3 \u2013 x2 = 1.0; Delta x4 = x4 \u2013 x3 = -1.75;\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Time <em>t<\/em><sub>i<\/sub> (min)<\/th>\r\n<th>Position $$ {x}_{i} $$ (km)<\/th>\r\n<th>Displacement $$ \\text{\u0394}{x}_{\\text{i}} $$ (km)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>$${t}_{0}=0$$<\/td>\r\n<td>$${x}_{0}=0$$<\/td>\r\n<td>$$\\text{\u0394}{x}_{0}=0$$<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>$${t}_{1}=9$$<\/td>\r\n<td>$${x}_{1}=0.5$$<\/td>\r\n<td>$$\\text{\u0394}{x}_{1}={x}_{1}-{x}_{0}=0.5$$<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>$${t}_{2}=18$$<\/td>\r\n<td>$${x}_{2}=0$$<\/td>\r\n<td>$$\\text{\u0394}{x}_{2}={x}_{2}-{x}_{1}=-0.5$$<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>$${t}_{3}=33$$<\/td>\r\n<td>$${x}_{3}=1.0$$<\/td>\r\n<td>$$\\text{\u0394}{x}_{3}={x}_{3}-{x}_{2}=1.0$$<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>$${t}_{4}=58$$<\/td>\r\n<td>$${x}_{4}=-0.75$$<\/td>\r\n<td>$$\\text{\u0394}{x}_{4}={x}_{4}-{x}_{3}=-1.75$$<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h4>Solution<\/h4>\r\n<ol id=\"fs-id1168329466395\" type=\"a\">\r\n \t<li>[reveal-answer q=\"905360\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"905360\"]From the above table, the total displacement is $$ \\sum \\text{\u0394}{x}_{\\text{i}}=0.5-0.5+1.0-1.75\\,\\text{km}=-0.75\\,\\text{km}\\text{.}$$[\/hidden-answer]<\/li>\r\n \t<li>[reveal-answer q=\"289407\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"289407\"]The magnitude of the total displacement is $$ |-0.75|\\,\\text{km}=0.75\\,\\text{km}$$.[\/hidden-answer]<\/li>\r\n \t<li>[reveal-answer q=\"368664\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"368664\"]$$\\text{Average velocity}=\\frac{\\text{Total}\\,\\text{displacement}}{\\text{Elapsed}\\,\\text{time}}=\\overset{\\text{\u2013}}{v}=\\frac{-0.75\\,\\text{km}}{58\\,\\text{min}}=-0.013\\,\\text{km\/min}$$[\/hidden-answer]<\/li>\r\n \t<li>[reveal-answer q=\"427772\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"427772\"]The total distance traveled (sum of magnitudes of individual displacements) is $$ {x}_{\\text{Total}}=\\sum |\\text{\u0394}{x}_{\\text{i}}|=0.5+0.5+1.0+1.75\\,\\text{km}=3.75\\,\\text{km}$$.[\/hidden-answer]<\/li>\r\n \t<li>[reveal-answer q=\"329442\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"329442\"]We can graph Jill\u2019s position versus time as a useful aid to see the motion; the graph is shown in (Figure).\r\n<div id=\"CNX_UPhysics_03_01_Pos\" class=\"wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"413\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31185009\/CNX_UPhysics_03_01_Pos.jpg\" alt=\"Graph shows position in kilometers plotted as a function of time in minutes.\" width=\"413\" height=\"395\" \/> <strong>Figure 3.5<\/strong> This graph depicts Jill\u2019s position versus time. The average velocity is the slope of a line connecting the initial and final points.[\/caption]\r\n\r\n<\/div>\r\n[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<h4>Significance<\/h4>\r\nJill\u2019s total displacement is \u22120.75 km, which means at the end of her trip she ends up $$ 0.75\\,\\text{km} $$ due west of her home. The average velocity means if someone was to walk due west at $$ 0.013 $$ km\/min starting at the same time Jill left her home, they both would arrive at the final stopping point at the same time. Note that if Jill were to end her trip at her house, her total displacement would be zero, as well as her average velocity. The total distance traveled during the 58 minutes of elapsed time for her trip is 3.75 km.\r\n\r\n<\/div>\r\n<div id=\"fs-id1168326766711\" class=\"textbox exercises check-understanding\">\r\n<h3>Check Your Understanding<\/h3>\r\n<div id=\"fs-id1168326766714\" class=\"problem textbox\">\r\n<div id=\"fs-id1168327092291\">\r\n<p id=\"fs-id1168327092293\">A cyclist rides 3 km west and then turns around and rides 2 km east. (a) What is his displacement? (b) What is the distance traveled? (c) What is the magnitude of his displacement?<\/p>\r\n<span id=\"fs-id1168326777142\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31185011\/CNX_UPhysics_03_01_CYU1_img.jpg\" alt=\"Figure shows timeline of cyclist\u2019s movement. First displacement is to the left by 3.0 kilometers. Second displacement is from the final point to the right by 2.0 kilometers.\" \/><\/span>\r\n\r\n<\/div>\r\n<div id=\"fs-id1168326836246\">\r\n<p id=\"fs-id1168326920106\">[reveal-answer q=\"151192\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"151192\"](a) The rider\u2019s displacement is $$ \\text{\u0394}x={x}_{\\text{f}}-{x}_{0}=-1\\,\\text{km}$$. (The displacement is negative because we take east to be positive and west to be negative.) (b) The distance traveled is 3 km + 2 km = 5 km. (c) The magnitude of the displacement is 1 km.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168326809070\" class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"fs-id1168327091282\">\r\n \t<li>Kinematics is the description of motion without considering its causes. In this chapter, it is limited to motion along a straight line, called one-dimensional motion.<\/li>\r\n \t<li>Displacement is the change in position of an object. The SI unit for displacement is the meter. Displacement has direction as well as magnitude.<\/li>\r\n \t<li>Distance traveled is the total length of the path traveled between two positions.<\/li>\r\n \t<li>Time is measured in terms of change. The time between two position points $$ {x}_{1} $$ and $$ {x}_{2} $$ is $$ \\text{\u0394}t={t}_{2}-{t}_{1}$$. Elapsed time for an event is $$ \\text{\u0394}t={t}_{\\text{f}}-{t}_{0}$$, where $$ {t}_{\\text{f}} $$ is the final time and $$ {t}_{0} $$ is the initial time. The initial time is often taken to be zero.<\/li>\r\n \t<li>Average velocity $$ \\overset{\\text{\u2013}}{v} $$ is defined as displacement divided by elapsed time. If $$ {x}_{1},{t}_{1} $$ and $$ {x}_{2},{t}_{2} $$ are two position time points, the average velocity between these points is\r\n<div id=\"fs-id1168329006610\" class=\"unnumbered\">$$\\overset{\\text{\u2013}}{v}=\\frac{\\text{\u0394}x}{\\text{\u0394}t}=\\frac{{x}_{2}-{x}_{1}}{{t}_{2}-{t}_{1}}.$$<\/div><\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1168329187572\" class=\"review-conceptual-questions\">\r\n<h3>Conceptual Questions<\/h3>\r\n<div id=\"fs-id1168327144875\" class=\"problem textbox\">\r\n<div id=\"fs-id1168327144877\">\r\n<p id=\"fs-id1168329137112\">Give an example in which there are clear distinctions among distance traveled, displacement, and magnitude of displacement. Identify each quantity in your example specifically.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1168329491939\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1168329491939\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1168329491939\"]\r\n<p id=\"fs-id1168326857976\">You drive your car into town and return to drive past your house to a friend\u2019s house.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168326781258\" class=\"problem textbox\">\r\n<div id=\"fs-id1168329516813\">\r\n<p id=\"fs-id1168329516815\">Under what circumstances does distance traveled equal magnitude of displacement? What is the only case in which magnitude of displacement and displacement are exactly the same?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168329042862\" class=\"problem textbox\">\r\n<div id=\"fs-id1168329042864\">\r\n<p id=\"fs-id1168329175042\">Bacteria move back and forth using their flagella (structures that look like little tails). Speeds of up to 50 \u03bcm\/s (50 \u00d7 10<sup>\u22126<\/sup> m\/s) have been observed. The total distance traveled by a bacterium is large for its size, whereas its displacement is small. Why is this?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1168326792665\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1168326792665\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1168326792665\"]\r\n<p id=\"fs-id1168329332700\">If the bacteria are moving back and forth, then the displacements are canceling each other and the final displacement is small.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168326851243\" class=\"problem textbox\">\r\n<div id=\"fs-id1168329050745\">\r\n<p id=\"fs-id1168329050747\">Give an example of a device used to measure time and identify what change in that device indicates a change in time.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168328991098\" class=\"problem textbox\">\r\n<div id=\"fs-id1168326786808\">\r\n<p id=\"fs-id1168326786810\">Does a car\u2019s odometer measure distance traveled or displacement?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1168329476976\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1168329476976\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1168329476976\"]\r\n<p id=\"fs-id1168329476978\">Distance traveled<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168329466274\" class=\"problem textbox\">\r\n<div id=\"fs-id1168329516953\">\r\n<p id=\"fs-id1168329516955\">During a given time interval the average velocity of an object is zero. What can you say conclude about its displacement over the time interval?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168329282324\" class=\"review-problems textbox exercises\">\r\n<h3>Problems<\/h3>\r\n<div id=\"fs-id1168329317598\" class=\"problem textbox\">\r\n<div id=\"fs-id1168329400845\">\r\n<p id=\"fs-id1168329400847\">Consider a coordinate system in which the positive <em>x<\/em> axis is directed upward vertically. What are the positions of a particle (a) 5.0 m directly above the origin and (b) 2.0 m below the origin?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168329490110\" class=\"problem textbox\">\r\n<div id=\"fs-id1168329490112\">\r\n<p id=\"fs-id1168329492788\">A car is 2.0 km west of a traffic light at <em>t<\/em> = 0 and 5.0 km east of the light at <em>t<\/em> = 6.0 min. Assume the origin of the coordinate system is the light and the positive <em>x<\/em> direction is eastward. (a) What are the car\u2019s position vectors at these two times? (b) What is the car\u2019s displacement between 0 min and 6.0 min?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1168329462686\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1168329462686\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1168329462686\"]\r\n<p id=\"fs-id1168329462688\">a. $$ {\\overset{\\to }{x}}_{1}=(-2.0\\,\\text{m})\\hat{i}$$, $$ {\\overset{\\to }{x}}_{2}=(5.0\\,\\text{m})\\hat{i}$$; b. 7.0 m east<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168326780779\" class=\"problem textbox\">\r\n<div id=\"fs-id1168326780782\">\r\n<p id=\"fs-id1168326860701\">The Shanghai maglev train connects Longyang Road to Pudong International Airport, a distance of 30 km. The journey takes 8 minutes on average. What is the maglev train\u2019s average velocity?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168326940079\" class=\"problem textbox\">\r\n<div id=\"fs-id1168326940081\">\r\n<p id=\"fs-id1168326805722\">The position of a particle moving along the <em>x<\/em>-axis is given by $$ x(t)=4.0-2.0t $$ m. (a) At what time does the particle cross the origin? (b) What is the displacement of the particle between $$ \\text{t}=3.0\\,\\text{s} $$ and $$ \\text{t}=6.0\\,\\text{s}?$$<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1168326770074\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1168326770074\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1168326770074\"]\r\n<p id=\"fs-id1168326770076\">a. $$ t=2.0 $$ s; b. $$ x(6.0)-x(3.0)=-8.0-(-2.0)=-6.0\\,\\text{m}$$<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168326798194\" class=\"problem textbox\">\r\n<div id=\"fs-id1168326822694\">\r\n<p id=\"fs-id1168326822697\">A cyclist rides 8.0 km east for 20 minutes, then he turns and heads west for 8 minutes and 3.2 km. Finally, he rides east for 16 km, which takes 40 minutes. (a) What is the final displacement of the cyclist? (b) What is his average velocity?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1168329063635\" class=\"problem textbox\">\r\n<div id=\"fs-id1168329063637\">\r\n<p id=\"fs-id1168329517547\">On February 15, 2013, a superbolide meteor (brighter than the Sun) entered Earth\u2019s atmosphere over Chelyabinsk, Russia, and exploded at an altitude of 23.5 km. Eyewitnesses could feel the intense heat from the fireball, and the blast wave from the explosion blew out windows in buildings. The blast wave took approximately 2 minutes 30 seconds to reach ground level. (a) What was the average velocity of the blast wave? b) Compare this with the speed of sound, which is 343 m\/s at sea level.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1168329517552\" class=\"solution\">\r\n\r\n[reveal-answer q=\"fs-id1168329517552\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"fs-id1168329517552\"]\r\n<p id=\"fs-id1168329318052\">a. 150.0 s, $$ \\overset{\\text{\u2013}}{v}=156.7\\,\\text{m\/s}$$; b. 45.7% the speed of sound at sea level<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Glossary<\/h3>\r\n<dl id=\"fs-id1168326791706\">\r\n \t<dt><strong>average velocity<\/strong><\/dt>\r\n \t<dd id=\"fs-id1168329400870\">the displacement divided by the time over which displacement occurs<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1168329400874\">\r\n \t<dt><strong>displacement<\/strong><\/dt>\r\n \t<dd id=\"fs-id1168329511515\">the change in position of an object<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1168326897289\">\r\n \t<dt><strong>distance traveled<\/strong><\/dt>\r\n \t<dd id=\"fs-id1168326925585\">the total length of the path traveled between two positions<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1168326925589\">\r\n \t<dt><strong>elapsed time<\/strong><\/dt>\r\n \t<dd id=\"fs-id1168326801364\">the difference between the ending time and the beginning time<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1168326794388\">\r\n \t<dt><strong>kinematics<\/strong><\/dt>\r\n \t<dd id=\"fs-id1168329394919\">the description of motion through properties such as position, time, velocity, and acceleration<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1168329394923\">\r\n \t<dt><strong>position<\/strong><\/dt>\r\n \t<dd id=\"fs-id1168329181740\">the location of an object at a particular time<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1168329478031\">\r\n \t<dt><strong>total displacement<\/strong><\/dt>\r\n \t<dd id=\"fs-id1168329193012\">the sum of individual displacements over a given time period<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Define position, displacement, and distance traveled.<\/li>\n<li>Calculate the total displacement given the position as a function of time.<\/li>\n<li>Determine the total distance traveled.<\/li>\n<li>Calculate the average velocity given the displacement and elapsed time.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1168329489850\">When you\u2019re in motion, the basic questions to ask are: Where are you? Where are you going? How fast are you getting there? The answers to these questions require that you specify your position, your displacement, and your average velocity\u2014the terms we define in this section.<\/p>\n<div id=\"fs-id1168329158454\" class=\"bc-section section\">\n<h3>Position<\/h3>\n<p id=\"fs-id1168326782871\">To describe the motion of an object, you must first be able to describe its<strong> position<\/strong> (<em>x<\/em>): <em>where it is at any particular time<\/em>. More precisely, we need to specify its position relative to a convenient frame of reference. A<strong> <span class=\"no-emphasis\">frame of reference<\/span><\/strong> is an arbitrary set of axes from which the position and motion of an object are described. Earth is often used as a frame of reference, and we often describe the position of an object as it relates to stationary objects on Earth. For example, a rocket launch could be described in terms of the position of the rocket with respect to Earth as a whole, whereas a cyclist\u2019s position could be described in terms of where she is in relation to the buildings she passes <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_03_01_Cyclists\">(Figure)<\/a>. In other cases, we use reference frames that are not stationary but are in motion relative to Earth. To describe the position of a person in an airplane, for example, we use the airplane, not Earth, as the reference frame. To describe the position of an object undergoing one-dimensional motion, we often use the variable <em>x<\/em>. Later in the chapter, during the discussion of free fall, we use the variable <em>y<\/em>.<\/p>\n<div id=\"CNX_UPhysics_03_01_Cyclists\" class=\"wp-caption aligncenter\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31184959\/CNX_UPhysics_03_01_Cyclists.jpg\" alt=\"Picture shows three people riding bicycles next to a canal.\" width=\"487\" height=\"366\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 3.2<\/strong> These cyclists in Vietnam can be described by their position relative to buildings or a canal. Their motion can be described by their change in position, or displacement, in a frame of reference. (credit: Suzan Black)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1168326761623\" class=\"bc-section section\">\n<h3>Displacement<\/h3>\n<p id=\"fs-id1168329187699\">If an object moves relative to a frame of reference\u2014for example, if a professor moves to the right relative to a whiteboard <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_03_01_Chalkboard\">(Figure)<\/a>\u2014then the object\u2019s position changes. This change in position is called <strong>displacement<\/strong>. The word <em>displacement<\/em> implies that an object has moved, or has been displaced. Although position is the numerical value of <em>x<\/em> along a straight line where an object might be located, displacement gives the <em>change<\/em> in position along this line. Since displacement indicates direction, it is a vector and can be either positive or negative, depending on the choice of positive direction. Also, an analysis of motion can have many displacements embedded in it. If right is positive and an object moves 2 m to the right, then 4 m to the left, the individual displacements are 2 m and $$ -4 $$ m, respectively.<\/p>\n<div id=\"CNX_UPhysics_03_01_Chalkboard\" class=\"wp-caption aligncenter\">\n<div style=\"width: 699px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31185003\/CNX_UPhysics_03_01_Chalkboard.jpg\" alt=\"Illustration shows professor at two different locations. The first location is marked as 1.5 meters at the x axis; the second location is marked as 3.5 meters at the x axis. The displacement between the two locations is 2 meters.\" width=\"689\" height=\"544\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 3.3<\/strong> A professor paces left and right while lecturing. Her position relative to Earth is given by x. The +2.0-m displacement of the professor relative to Earth is represented by an arrow pointing to the right.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1168328983002\">\n<div><strong>Displacement<\/strong><\/div>\n<p id=\"fs-id1168327144862\">Displacement $$ \\text{\u0394}x $$ is the change in position of an object:<\/p>\n<div id=\"fs-id1168329338892\">$$\\text{\u0394}x={x}_{\\text{f}}-{x}_{0},$$<\/div>\n<p id=\"fs-id1168326911738\">where $$ \\text{\u0394}x $$ is displacement, $$ {x}_{\\text{f}} $$ is the final position, and $$ {x}_{0} $$ is the initial position.<\/p>\n<\/div>\n<p id=\"fs-id1168326795409\">We use the uppercase Greek letter delta (\u0394) to mean \u201cchange in\u201d whatever quantity follows it; thus, $$ \\text{\u0394}x $$ means <em>change in position<\/em> (final position less initial position). We always solve for displacement by subtracting initial position $$ {x}_{0} $$ from final position $$ {x}_{\\text{f}}$$. Note that the SI unit for displacement is the meter, but sometimes we use kilometers or other units of length. Keep in mind that when units other than meters are used in a problem, you may need to convert them to meters to complete the calculation (see <a href=\"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/back-matter\/conversion-factors\/\">Conversion Factors<\/a>).<\/p>\n<p id=\"fs-id1168329464556\">Objects in motion can also have a series of displacements. In the previous example of the pacing professor, the individual displacements are 2 m and $$ -4 $$ m, giving a total displacement of \u22122 m. We define<strong> total displacement<\/strong> $$ \\text{\u0394}{x}_{\\text{Total}}$$, as <em>the sum of the individual displacements<\/em>, and express this mathematically with the equation<\/p>\n<div class=\"equation-callout\">\n<div id=\"fs-id1168326821763\">$$\\text{\u0394}{x}_{\\text{Total}}=\\sum \\text{\u0394}{x}_{\\text{i}},$$<\/div>\n<\/div>\n<p id=\"fs-id1168329069172\">where $$ \\text{\u0394}{x}_{i} $$ are the individual displacements. In the earlier example,<\/p>\n<div id=\"fs-id1168326796770\" class=\"unnumbered\">$$\\text{\u0394}{x}_{1}={x}_{1}-{x}_{0}=2-0=2\\,\\text{m.}$$<\/div>\n<p id=\"fs-id1168329410562\">Similarly,<\/p>\n<div id=\"fs-id1168329155289\" class=\"unnumbered\">$$\\text{\u0394}{x}_{2}={x}_{2}-{x}_{1}=-2-(2)=-4\\,\\text{m.}$$<\/div>\n<p id=\"fs-id1168329081791\">Thus,<\/p>\n<div id=\"fs-id1168326753395\" class=\"unnumbered\">$$\\text{\u0394}{x}_{\\text{Total}}=\\text{\u0394}{x}_{1}+\\text{\u0394}{x}_{2}=2-4=-2\\,\\text{m}\\text{\u200b.}$$<\/div>\n<p id=\"fs-id1168329518900\">The total displacement is 2 \u2212 4 = \u22122 m to the left, or in the negative direction. It is also useful to calculate the magnitude of the displacement, or its size. The magnitude of the displacement is always positive. This is the absolute value of the displacement, because displacement is a vector and cannot have a negative value of magnitude. In our example, the magnitude of the total displacement is 2 m, whereas the magnitudes of the individual displacements are 2 m and 4 m.<\/p>\n<p id=\"fs-id1168326908350\">The magnitude of the total displacement should not be confused with the distance traveled. Distance traveled $$ {x}_{\\text{Total}}$$, is the total length of the path traveled between two positions. In the previous problem, the <strong>distance traveled<\/strong> is the sum of the magnitudes of the individual displacements:<\/p>\n<div id=\"fs-id1168329266306\" class=\"unnumbered\">$${x}_{\\text{Total}}=|\\text{\u0394}{x}_{1}|+|\\text{\u0394}{x}_{2}|=2+4=6\\,\\text{m}\\text{.}$$<\/div>\n<\/div>\n<div id=\"fs-id1168326783743\" class=\"bc-section section\">\n<h3>Average Velocity<\/h3>\n<p id=\"fs-id1168329046092\">To calculate the other physical quantities in kinematics we must introduce the time variable. The time variable allows us not only to state where the object is (its position) during its motion, but also how fast it is moving. How fast an object is moving is given by the rate at which the position changes with time.<\/p>\n<p id=\"fs-id1168329333293\">For each position $$ {x}_{\\text{i}}$$, we assign a particular time $$ {t}_{\\text{i}}$$. If the details of the motion at each instant are not important, the rate is usually expressed as the <strong>average velocity<\/strong> $$ \\overset{\\text{\u2013}}{v}$$. This vector quantity is simply the total displacement between two points divided by the time taken to travel between them. The time taken to travel between two points is called the <strong>elapsed time<\/strong> $$ \\text{\u0394}t$$.<\/p>\n<div id=\"fs-id1168329050548\">\n<div>Average Velocity<\/div>\n<p id=\"fs-id1168329192944\">If $$ {x}_{1} $$ and $$ {x}_{2} $$ are the positions of an object at times $$ {t}_{1} $$ and $$ {t}_{2}$$, respectively, then<\/p>\n<div id=\"fs-id1168326922062\">$$\\begin{array}{cc} \\text{Average velocity}=\\overset{\\text{\u2013}}{v}=\\frac{\\text{Displacement between two points}}{\\text{Elapsed time between two points}}\\\\ \\overset{\\text{\u2013}}{v}=\\frac{\\text{\u0394}x}{\\text{\u0394}t}=\\frac{{x}_{2}-{x}_{1}}{{t}_{2}-{t}_{1}}.\\end{array}$$<\/div>\n<\/div>\n<p id=\"fs-id1168326875739\">It is important to note that the average velocity is a vector and can be negative, depending on positions $$ {x}_{1} $$ and $$ {x}_{2}$$.<\/p>\n<div id=\"fs-id1168329011730\" class=\"textbox examples\">\n<h3>Example<\/h3>\n<h4>Delivering Flyers<\/h4>\n<p>Jill sets out from her home to deliver flyers for her yard sale, traveling due east along her street lined with houses. At $$ 0.5 $$ km and 9 minutes later she runs out of flyers and has to retrace her steps back to her house to get more. This takes an additional 9 minutes. After picking up more flyers, she sets out again on the same path, continuing where she left off, and ends up 1.0 km from her house. This third leg of her trip takes $$ 15 $$ minutes. At this point she turns back toward her house, heading west. After $$ 1.75 $$ km and $$ 25 $$ minutes she stops to rest.<\/p>\n<ol id=\"fs-id1168329101865\" type=\"a\">\n<li>What is Jill\u2019s total displacement to the point where she stops to rest?<\/li>\n<li>What is the magnitude of the final displacement?<\/li>\n<li>What is the average velocity during her entire trip?<\/li>\n<li>What is the total distance traveled?<\/li>\n<li>Make a graph of position versus time.<\/li>\n<\/ol>\n<p id=\"fs-id1168329133729\">A sketch of Jill\u2019s movements is shown in <a class=\"autogenerated-content\" href=\"#CNX_UPhysics_03_01_Flyers\">(Figure)<\/a>.<\/p>\n<div id=\"CNX_UPhysics_03_01_Flyers\" class=\"wp-caption aligncenter\">\n<div style=\"width: 795px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31185006\/CNX_UPhysics_03_01_Flyers.jpg\" alt=\"Figure shows a timeline of a person\u2019s movement. First displacement is from the home to the right by 0.5 kilometers. Second displacement is back to the starting point. Third displacement is to the right by 1.0 kilometer. Fourth displacement is from the final point to the left by 1.75 kilometers.\" width=\"785\" height=\"333\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3.4 Timeline of Jill\u2019s movements.<\/p>\n<\/div>\n<\/div>\n<h4>Strategy<\/h4>\n<p>The problem contains data on the various legs of Jill\u2019s trip, so it would be useful to make a table of the physical quantities. We are given position and time in the wording of the problem so we can calculate the displacements and the elapsed time. We take east to be the positive direction. From this information we can find the total displacement and average velocity. Jill\u2019s home is the starting point $$ {x}_{0}$$. The following table gives Jill\u2019s time and position in the first two columns, and the displacements are calculated in the third column.<\/p>\n<table id=\"fs-id1168326821006\" class=\"unnumbered\" summary=\"This table has three columns and six rows. The first row is a header row and it labels each column: \u201cTime ti in minutes\u201d, \u201cPosition xi in kilometers\u201d, \u201cDisplacement Delta xi in kilometers\u201d. Under the \u201cTime ti in minutes\u201d: t0 = 0; t1 = 9; t2 = 18; t3 = 33; t4 = 58; Under the \u201cPosition xi in kilometers\u201d: x0 = 0; x1 = 0.5; x2 = 0; x3 = 1.0; x4 = -0.75; Under the \u201cDisplacement Delta xi in kilometers\u201d: Delta x0 = 0; Delta x1 = x1 \u2013 x0 = 0.5; Delta x2 = x2 \u2013 x1 = -0.5; Delta x3 = x3 \u2013 x2 = 1.0; Delta x4 = x4 \u2013 x3 = -1.75;\">\n<thead>\n<tr valign=\"top\">\n<th>Time <em>t<\/em><sub>i<\/sub> (min)<\/th>\n<th>Position $$ {x}_{i} $$ (km)<\/th>\n<th>Displacement $$ \\text{\u0394}{x}_{\\text{i}} $$ (km)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>$${t}_{0}=0$$<\/td>\n<td>$${x}_{0}=0$$<\/td>\n<td>$$\\text{\u0394}{x}_{0}=0$$<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>$${t}_{1}=9$$<\/td>\n<td>$${x}_{1}=0.5$$<\/td>\n<td>$$\\text{\u0394}{x}_{1}={x}_{1}-{x}_{0}=0.5$$<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>$${t}_{2}=18$$<\/td>\n<td>$${x}_{2}=0$$<\/td>\n<td>$$\\text{\u0394}{x}_{2}={x}_{2}-{x}_{1}=-0.5$$<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>$${t}_{3}=33$$<\/td>\n<td>$${x}_{3}=1.0$$<\/td>\n<td>$$\\text{\u0394}{x}_{3}={x}_{3}-{x}_{2}=1.0$$<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>$${t}_{4}=58$$<\/td>\n<td>$${x}_{4}=-0.75$$<\/td>\n<td>$$\\text{\u0394}{x}_{4}={x}_{4}-{x}_{3}=-1.75$$<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4>Solution<\/h4>\n<ol id=\"fs-id1168329466395\" type=\"a\">\n<li>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q905360\">Show Answer<\/span><\/p>\n<div id=\"q905360\" class=\"hidden-answer\" style=\"display: none\">From the above table, the total displacement is $$ \\sum \\text{\u0394}{x}_{\\text{i}}=0.5-0.5+1.0-1.75\\,\\text{km}=-0.75\\,\\text{km}\\text{.}$$<\/div>\n<\/div>\n<\/li>\n<li>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q289407\">Show Answer<\/span><\/p>\n<div id=\"q289407\" class=\"hidden-answer\" style=\"display: none\">The magnitude of the total displacement is $$ |-0.75|\\,\\text{km}=0.75\\,\\text{km}$$.<\/div>\n<\/div>\n<\/li>\n<li>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q368664\">Show Answer<\/span><\/p>\n<div id=\"q368664\" class=\"hidden-answer\" style=\"display: none\">$$\\text{Average velocity}=\\frac{\\text{Total}\\,\\text{displacement}}{\\text{Elapsed}\\,\\text{time}}=\\overset{\\text{\u2013}}{v}=\\frac{-0.75\\,\\text{km}}{58\\,\\text{min}}=-0.013\\,\\text{km\/min}$$<\/div>\n<\/div>\n<\/li>\n<li>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q427772\">Show Answer<\/span><\/p>\n<div id=\"q427772\" class=\"hidden-answer\" style=\"display: none\">The total distance traveled (sum of magnitudes of individual displacements) is $$ {x}_{\\text{Total}}=\\sum |\\text{\u0394}{x}_{\\text{i}}|=0.5+0.5+1.0+1.75\\,\\text{km}=3.75\\,\\text{km}$$.<\/div>\n<\/div>\n<\/li>\n<li>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q329442\">Show Answer<\/span><\/p>\n<div id=\"q329442\" class=\"hidden-answer\" style=\"display: none\">We can graph Jill\u2019s position versus time as a useful aid to see the motion; the graph is shown in (Figure).<\/p>\n<div id=\"CNX_UPhysics_03_01_Pos\" class=\"wp-caption aligncenter\">\n<div style=\"width: 423px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31185009\/CNX_UPhysics_03_01_Pos.jpg\" alt=\"Graph shows position in kilometers plotted as a function of time in minutes.\" width=\"413\" height=\"395\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 3.5<\/strong> This graph depicts Jill\u2019s position versus time. The average velocity is the slope of a line connecting the initial and final points.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n<\/ol>\n<h4>Significance<\/h4>\n<p>Jill\u2019s total displacement is \u22120.75 km, which means at the end of her trip she ends up $$ 0.75\\,\\text{km} $$ due west of her home. The average velocity means if someone was to walk due west at $$ 0.013 $$ km\/min starting at the same time Jill left her home, they both would arrive at the final stopping point at the same time. Note that if Jill were to end her trip at her house, her total displacement would be zero, as well as her average velocity. The total distance traveled during the 58 minutes of elapsed time for her trip is 3.75 km.<\/p>\n<\/div>\n<div id=\"fs-id1168326766711\" class=\"textbox exercises check-understanding\">\n<h3>Check Your Understanding<\/h3>\n<div id=\"fs-id1168326766714\" class=\"problem textbox\">\n<div id=\"fs-id1168327092291\">\n<p id=\"fs-id1168327092293\">A cyclist rides 3 km west and then turns around and rides 2 km east. (a) What is his displacement? (b) What is the distance traveled? (c) What is the magnitude of his displacement?<\/p>\n<p><span id=\"fs-id1168326777142\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2952\/2018\/01\/31185011\/CNX_UPhysics_03_01_CYU1_img.jpg\" alt=\"Figure shows timeline of cyclist\u2019s movement. First displacement is to the left by 3.0 kilometers. Second displacement is from the final point to the right by 2.0 kilometers.\" \/><\/span><\/p>\n<\/div>\n<div id=\"fs-id1168326836246\">\n<p id=\"fs-id1168326920106\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q151192\">Show Answer<\/span><\/p>\n<div id=\"q151192\" class=\"hidden-answer\" style=\"display: none\">(a) The rider\u2019s displacement is $$ \\text{\u0394}x={x}_{\\text{f}}-{x}_{0}=-1\\,\\text{km}$$. (The displacement is negative because we take east to be positive and west to be negative.) (b) The distance traveled is 3 km + 2 km = 5 km. (c) The magnitude of the displacement is 1 km.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1168326809070\" class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<ul id=\"fs-id1168327091282\">\n<li>Kinematics is the description of motion without considering its causes. In this chapter, it is limited to motion along a straight line, called one-dimensional motion.<\/li>\n<li>Displacement is the change in position of an object. The SI unit for displacement is the meter. Displacement has direction as well as magnitude.<\/li>\n<li>Distance traveled is the total length of the path traveled between two positions.<\/li>\n<li>Time is measured in terms of change. The time between two position points $$ {x}_{1} $$ and $$ {x}_{2} $$ is $$ \\text{\u0394}t={t}_{2}-{t}_{1}$$. Elapsed time for an event is $$ \\text{\u0394}t={t}_{\\text{f}}-{t}_{0}$$, where $$ {t}_{\\text{f}} $$ is the final time and $$ {t}_{0} $$ is the initial time. The initial time is often taken to be zero.<\/li>\n<li>Average velocity $$ \\overset{\\text{\u2013}}{v} $$ is defined as displacement divided by elapsed time. If $$ {x}_{1},{t}_{1} $$ and $$ {x}_{2},{t}_{2} $$ are two position time points, the average velocity between these points is\n<div id=\"fs-id1168329006610\" class=\"unnumbered\">$$\\overset{\\text{\u2013}}{v}=\\frac{\\text{\u0394}x}{\\text{\u0394}t}=\\frac{{x}_{2}-{x}_{1}}{{t}_{2}-{t}_{1}}.$$<\/div>\n<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1168329187572\" class=\"review-conceptual-questions\">\n<h3>Conceptual Questions<\/h3>\n<div id=\"fs-id1168327144875\" class=\"problem textbox\">\n<div id=\"fs-id1168327144877\">\n<p id=\"fs-id1168329137112\">Give an example in which there are clear distinctions among distance traveled, displacement, and magnitude of displacement. Identify each quantity in your example specifically.<\/p>\n<\/div>\n<div id=\"fs-id1168329491939\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1168329491939\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1168329491939\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1168326857976\">You drive your car into town and return to drive past your house to a friend\u2019s house.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1168326781258\" class=\"problem textbox\">\n<div id=\"fs-id1168329516813\">\n<p id=\"fs-id1168329516815\">Under what circumstances does distance traveled equal magnitude of displacement? What is the only case in which magnitude of displacement and displacement are exactly the same?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1168329042862\" class=\"problem textbox\">\n<div id=\"fs-id1168329042864\">\n<p id=\"fs-id1168329175042\">Bacteria move back and forth using their flagella (structures that look like little tails). Speeds of up to 50 \u03bcm\/s (50 \u00d7 10<sup>\u22126<\/sup> m\/s) have been observed. The total distance traveled by a bacterium is large for its size, whereas its displacement is small. Why is this?<\/p>\n<\/div>\n<div id=\"fs-id1168326792665\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1168326792665\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1168326792665\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1168329332700\">If the bacteria are moving back and forth, then the displacements are canceling each other and the final displacement is small.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1168326851243\" class=\"problem textbox\">\n<div id=\"fs-id1168329050745\">\n<p id=\"fs-id1168329050747\">Give an example of a device used to measure time and identify what change in that device indicates a change in time.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1168328991098\" class=\"problem textbox\">\n<div id=\"fs-id1168326786808\">\n<p id=\"fs-id1168326786810\">Does a car\u2019s odometer measure distance traveled or displacement?<\/p>\n<\/div>\n<div id=\"fs-id1168329476976\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1168329476976\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1168329476976\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1168329476978\">Distance traveled<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1168329466274\" class=\"problem textbox\">\n<div id=\"fs-id1168329516953\">\n<p id=\"fs-id1168329516955\">During a given time interval the average velocity of an object is zero. What can you say conclude about its displacement over the time interval?<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1168329282324\" class=\"review-problems textbox exercises\">\n<h3>Problems<\/h3>\n<div id=\"fs-id1168329317598\" class=\"problem textbox\">\n<div id=\"fs-id1168329400845\">\n<p id=\"fs-id1168329400847\">Consider a coordinate system in which the positive <em>x<\/em> axis is directed upward vertically. What are the positions of a particle (a) 5.0 m directly above the origin and (b) 2.0 m below the origin?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1168329490110\" class=\"problem textbox\">\n<div id=\"fs-id1168329490112\">\n<p id=\"fs-id1168329492788\">A car is 2.0 km west of a traffic light at <em>t<\/em> = 0 and 5.0 km east of the light at <em>t<\/em> = 6.0 min. Assume the origin of the coordinate system is the light and the positive <em>x<\/em> direction is eastward. (a) What are the car\u2019s position vectors at these two times? (b) What is the car\u2019s displacement between 0 min and 6.0 min?<\/p>\n<\/div>\n<div id=\"fs-id1168329462686\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1168329462686\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1168329462686\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1168329462688\">a. $$ {\\overset{\\to }{x}}_{1}=(-2.0\\,\\text{m})\\hat{i}$$, $$ {\\overset{\\to }{x}}_{2}=(5.0\\,\\text{m})\\hat{i}$$; b. 7.0 m east<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1168326780779\" class=\"problem textbox\">\n<div id=\"fs-id1168326780782\">\n<p id=\"fs-id1168326860701\">The Shanghai maglev train connects Longyang Road to Pudong International Airport, a distance of 30 km. The journey takes 8 minutes on average. What is the maglev train\u2019s average velocity?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1168326940079\" class=\"problem textbox\">\n<div id=\"fs-id1168326940081\">\n<p id=\"fs-id1168326805722\">The position of a particle moving along the <em>x<\/em>-axis is given by $$ x(t)=4.0-2.0t $$ m. (a) At what time does the particle cross the origin? (b) What is the displacement of the particle between $$ \\text{t}=3.0\\,\\text{s} $$ and $$ \\text{t}=6.0\\,\\text{s}?$$<\/p>\n<\/div>\n<div id=\"fs-id1168326770074\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1168326770074\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1168326770074\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1168326770076\">a. $$ t=2.0 $$ s; b. $$ x(6.0)-x(3.0)=-8.0-(-2.0)=-6.0\\,\\text{m}$$<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1168326798194\" class=\"problem textbox\">\n<div id=\"fs-id1168326822694\">\n<p id=\"fs-id1168326822697\">A cyclist rides 8.0 km east for 20 minutes, then he turns and heads west for 8 minutes and 3.2 km. Finally, he rides east for 16 km, which takes 40 minutes. (a) What is the final displacement of the cyclist? (b) What is his average velocity?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1168329063635\" class=\"problem textbox\">\n<div id=\"fs-id1168329063637\">\n<p id=\"fs-id1168329517547\">On February 15, 2013, a superbolide meteor (brighter than the Sun) entered Earth\u2019s atmosphere over Chelyabinsk, Russia, and exploded at an altitude of 23.5 km. Eyewitnesses could feel the intense heat from the fireball, and the blast wave from the explosion blew out windows in buildings. The blast wave took approximately 2 minutes 30 seconds to reach ground level. (a) What was the average velocity of the blast wave? b) Compare this with the speed of sound, which is 343 m\/s at sea level.<\/p>\n<\/div>\n<div id=\"fs-id1168329517552\" class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1168329517552\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1168329517552\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1168329318052\">a. 150.0 s, $$ \\overset{\\text{\u2013}}{v}=156.7\\,\\text{m\/s}$$; b. 45.7% the speed of sound at sea level<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl id=\"fs-id1168326791706\">\n<dt><strong>average velocity<\/strong><\/dt>\n<dd id=\"fs-id1168329400870\">the displacement divided by the time over which displacement occurs<\/dd>\n<\/dl>\n<dl id=\"fs-id1168329400874\">\n<dt><strong>displacement<\/strong><\/dt>\n<dd id=\"fs-id1168329511515\">the change in position of an object<\/dd>\n<\/dl>\n<dl id=\"fs-id1168326897289\">\n<dt><strong>distance traveled<\/strong><\/dt>\n<dd id=\"fs-id1168326925585\">the total length of the path traveled between two positions<\/dd>\n<\/dl>\n<dl id=\"fs-id1168326925589\">\n<dt><strong>elapsed time<\/strong><\/dt>\n<dd id=\"fs-id1168326801364\">the difference between the ending time and the beginning time<\/dd>\n<\/dl>\n<dl id=\"fs-id1168326794388\">\n<dt><strong>kinematics<\/strong><\/dt>\n<dd id=\"fs-id1168329394919\">the description of motion through properties such as position, time, velocity, and acceleration<\/dd>\n<\/dl>\n<dl id=\"fs-id1168329394923\">\n<dt><strong>position<\/strong><\/dt>\n<dd id=\"fs-id1168329181740\">the location of an object at a particular time<\/dd>\n<\/dl>\n<dl id=\"fs-id1168329478031\">\n<dt><strong>total displacement<\/strong><\/dt>\n<dd id=\"fs-id1168329193012\">the sum of individual displacements over a given time period<\/dd>\n<\/dl>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-359\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>OpenStax University Physics. <strong>Authored by<\/strong>: OpenStax CNX. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/cnx.org\/contents\/1Q9uMg_a@10.16:Gofkr9Oy@15\">https:\/\/cnx.org\/contents\/1Q9uMg_a@10.16:Gofkr9Oy@15<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/d50f6e32-0fda-46ef-a362-9bd36ca7c97d@10.16<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t 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http:\/\/cnx.org\/contents\/d50f6e32-0fda-46ef-a362-9bd36ca7c97d@10.16\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":"all-rights-reserved"},"chapter-type":[],"contributor":[],"license":[56],"class_list":["post-359","chapter","type-chapter","status-publish","hentry","license-all-rights-reserved"],"part":309,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters\/359","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/users\/311"}],"version-history":[{"count":11,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters\/359\/revisions"}],"predecessor-version":[{"id":2213,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters\/359\/revisions\/2213"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/parts\/309"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapters\/359\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/media?parent=359"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/pressbooks\/v2\/chapter-type?post=359"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/contributor?post=359"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/wp-json\/wp\/v2\/license?post=359"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}