{"id":77,"date":"2017-09-18T18:08:41","date_gmt":"2017-09-18T22:08:41","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/chapter\/2-1-displacement\/"},"modified":"2021-05-09T16:18:32","modified_gmt":"2021-05-09T20:18:32","slug":"2-1-displacement","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/chapter\/2-1-displacement\/","title":{"raw":"2.1 Displacement","rendered":"2.1 Displacement"},"content":{"raw":"<figure id=\"import-auto-id2723149\"><figcaption><\/figcaption>\n<div class=\"bcc-box bcc-highlight\">\n<h3>Summary<\/h3>\n<ul>\n \t<li>Define position, displacement, distance, and distance traveled.<\/li>\n \t<li>Explain the relationship between position and displacement.<\/li>\n \t<li>Distinguish between displacement and distance traveled.<\/li>\n \t<li>Calculate displacement and distance given initial position, final position and the path between the two.<\/li>\n<\/ul>\n<\/div>\n\n[caption id=\"\" align=\"aligncenter\" width=\"365\"]<img src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104\/wp-content\/uploads\/sites\/1393\/2017\/06\/Figure_02_01_00-1.jpg\" alt=\"Three people cycling along a canal. The blurred buildings in the background convey a sense of motion of the cyclists.\" width=\"365\" height=\"274\"> <strong>Figure 1.<\/strong> These cyclists in Vietnam can be described by their position relative to buildings and a canal. Their motion can be described by their change in position, or displacement, in the frame of reference. (credit: Suzan Black, Fotopedia).[\/caption]<\/figure>\n<section id=\"fs-id3178358\">\n<h1>Position<\/h1>\n<p id=\"import-auto-id1572700\">In order to describe the motion of an object, you must first be able to describe its <strong>position<\/strong>\u2014where it is at any particular time. More precisely, you need to specify its position relative to a convenient reference frame. Earth is often used as a reference frame, and we often describe the position of an object as it relates to stationary objects in that reference frame. For example, a rocket launch would be described in terms of the position of the rocket with respect to the Earth as a whole, while a professor\u2019s position could be described in terms of where she is in relation to the nearby white board. \u00a0In other cases, we use reference frames that are not stationary but are in motion relative to the Earth. To describe the position of a person in an airplane, for example, we use the airplane, not the Earth, as the reference frame. Please see the figures below.<\/p>\n\n<\/section><section id=\"fs-id2572460\">\n<h1>Displacement<\/h1>\n<p id=\"import-auto-id3310684\">If an object moves relative to a reference frame (for example, if a professor moves to the right relative to a white board or a passenger moves toward the rear of an airplane), then the object\u2019s position changes. This change in position is known as <strong>displacement<\/strong>. The word \u201cdisplacement\u201d implies that an object has moved, or has been displaced.<\/p>\n\n<div id=\"fs-id3206000\" class=\"note\">\n<div class=\"textbox shaded\">\n<h3>DISPLACEMENT<\/h3>\n<p id=\"import-auto-id2986917\">Displacement is the <em>change in position<\/em> of an object:<\/p>\n\n<div id=\"eip-458\" class=\"equation\" style=\"text-align: center\"><strong>\u0394<em>x =x<sub>f\u00a0\u00a0<\/sub>\u00a0-\u00a0x<sub>0<\/sub><\/em><\/strong><\/div>\n<div class=\"equation\" style=\"text-align: center\"><\/div>\n<p id=\"import-auto-id2795085\">where\u00a0<strong>\u0394<em>x<\/em><\/strong> is displacement, <em><strong>x<sub>f<\/sub><\/strong><\/em> is the final position, and <strong><em>x<\/em><sub>0<\/sub><\/strong> is the initial position.<\/p>\n\n<\/div>\n<p id=\"import-auto-id2804940\">In this text the upper case Greek letter\u00a0<strong>\u0394<\/strong> (delta) always means \u201cchange in\u201d whatever quantity follows it; thus, <strong>\u0394<em>x<\/em><\/strong> means <em>change in position<\/em>. Always solve for displacement by subtracting initial position <strong><em>x<\/em><sub>0<\/sub><\/strong> from final <em><strong>x<sub>f<\/sub><\/strong><\/em>.<em>\n<\/em><\/p>\n<p id=\"import-auto-id2880746\">Note that the SI unit for displacement is the meter (m) (see <a href=\"\/douglasphys1104summer2021\/part\/chapter-1-the-nature-of-science-and-physics\/\">Chapter 1 Physical Quantities and Units<\/a>), but sometimes kilometers, miles, feet, and other units of length are used. Keep in mind that when units other than the meter are used in a problem, you may need to convert them into meters to complete the calculation.<\/p>\n\n<\/div>\n\n[caption id=\"attachment_4397\" align=\"aligncenter\" width=\"705\"]<a href=\"\/douglasphys1104summer2021\/chapter\/2-1-displacement\/figure_02_01_01-1-2\/\"><img class=\"wp-image-73 size-full\" src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_02_01_01-1.jpg\" alt=\"Figure 2. Professor walking in front of a chalkboard. \" width=\"705\" height=\"560\"><\/a> <a id=\"\u201dFigure\" href=\"\"><\/a>Figure 2. 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]\n\n<\/section>&nbsp;\n\n<section id=\"fs-id2572460\">\n<div id=\"fs-id3206000\" class=\"note\">\n<figure id=\"import-auto-id2707699\">\n\n[caption id=\"\" align=\"aligncenter\" width=\"541\"]<a href=\"\/wp-content\/uploads\/sites\/153\/2017\/06\/Figure_02_01_02-1.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_02_01_02-1.jpg\" alt=\"View of an airplane with an inset of the passengers sitting inside. A passenger has just moved from his seat and is now standing in the back. His initial position was 6 point 0 meters. His final position is 2 point 0 meters. His displacement is given by the equation delta x equals x sub f minus x sub 0 equals 4 point zero meters.\" width=\"541\" height=\"399\"><\/a> <strong>Figure 3.<\/strong> A passenger moves from his seat to the back of the plane. His location relative to the airplane is given by <em><strong>x<\/strong><\/em>. The <strong>-4.0-m<\/strong> displacement of the passenger relative to the plane is represented by an arrow toward the rear of the plane. Notice that the arrow representing his displacement is twice as long as the arrow representing the displacement of the professor shown above (he moves twice as far).[\/caption]<\/figure>\n<p id=\"import-auto-id2815802\">Note that displacement has a direction as well as a magnitude. The professor\u2019s displacement is 2.0 m to the right, and the airline passenger\u2019s displacement is 4.0 m toward the rear. In one-dimensional motion, direction can be specified with a plus or minus sign. When you begin a problem, you should select which direction is positive (usually that will be to the right or up, but you are free to select positive as being any direction). The professor\u2019s initial position is <strong><em>x<\/em><sub>0<\/sub> = 1.5 m<\/strong> and her final position is <strong><em>x<sub>f<\/sub><\/em> = 3.5 m<\/strong>. Thus her displacement is<\/p>\n\n<div id=\"eip-556\" class=\"equation\" style=\"text-align: center\"><strong>\u00a0\u0394x = x <sub>final<\/sub>\u00a0 -x <sub>initial<\/sub> = x<sub>f<\/sub> - x<sub>o<\/sub><\/strong> = 3.5 m -1.5 m\u00a0 = +2.0 m<\/div>\n<p id=\"import-auto-id1373168\">In this coordinate system, motion to the right is positive, whereas motion to the left is negative. Similarly, the airplane passenger\u2019s initial position is <strong><em>x<\/em><sub>0<\/sub> = 6.0 m<\/strong> and his final position is <strong><em>x<sub>f<\/sub><\/em> = 2.0 m<\/strong>, so his displacement is<\/p>\n\n<\/div>\n<p style=\"text-align: center\"><strong>\u00a0\u0394x = x <sub>final<\/sub>\u00a0 -x <sub>initial<\/sub> = x<sub>f<\/sub> - x<sub>o<\/sub><\/strong> = 2.0 m - 6.0\u00a0 m\u00a0 = -4.0 m<\/p>\n\n<div id=\"fs-id3206000\" class=\"note\">\n<p id=\"import-auto-id3579333\">His displacement is negative because his motion is toward the rear of the plane, or in the negative x direction in our coordinate system.<\/p>\n\n<\/div>\n<\/section><section id=\"fs-id1414454\">\n<h1>Distance<\/h1>\n<p id=\"import-auto-id3580245\">Although displacement is described in terms of direction, distance is not. <strong>Distance <\/strong>is defined to be <em>the magnitude or size of displacement between two positions<\/em>. Note that the distance between two positions is not the same as the distance traveled between them. <strong>Distance traveled<\/strong> is <em>the total length of the path traveled between two positions<\/em>. Distance has no direction and, thus, no sign. For example, the distance the professor walks is 2.0 m. The distance the airplane passenger walks is 4.0 m.<\/p>\n\n<div id=\"fs-id3044492\" class=\"note\">\n<div class=\"textbox shaded\">\n<h3>MISCONCEPTION ALERT: DISTANCE TRAVELED VS. MAGNITUDE OF DISPLACEMENT<\/h3>\n<p id=\"import-auto-id2850966\">It is important to note that the <em>distance traveled<\/em>, however, can be greater than the magnitude of the displacement (by magnitude, we mean just the size of the displacement without regard to its direction; that is, just a number with a unit). For example, the professor could pace back and forth many times, perhaps walking a distance of 150 m during a lecture, yet still end up only 2.0 m to the right of her starting point. In this case her displacement would be +2.0 m, the magnitude of her displacement would be 2.0 m, but the distance she traveled would be 150 m. In kinematics we nearly always deal with displacement and magnitude of displacement, and almost never with distance traveled. One way to think about this is to assume you marked the start of the motion and the end of the motion. The displacement is simply the difference in the position of the two marks and is independent of the path taken in traveling between the two marks. The distance traveled, however, is the total length of the path taken between the two marks.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id3589986\" class=\"exercise\">\n<div class=\"bcc-box bcc-info\">\n<h3>Check Your Understanding 1<\/h3>\n<div id=\"fs-id2996632\" class=\"problem\">\n<p id=\"import-auto-id3225677\"><strong>1:<\/strong> A cyclist rides 3 km west and then turns around and rides 2 km east. (a) What is her displacement? (b) What distance does she ride? (c) What is the magnitude of her displacement?<\/p>\n\n<\/div>\n<\/div>\n<h1>\u00a0Section Summary<\/h1>\n<ul>\n \t<li>Kinematics is the study of motion without considering its causes. In this chapter, it is limited to motion along a straight line, called one-dimensional motion.<\/li>\n \t<li>Displacement is the change in position of an object.<\/li>\n \t<li>In symbols, displacement\u00a0<strong>\u0394<em>x<\/em><\/strong> is defined to be<\/li>\n<\/ul>\n<p style=\"text-align: center\"><strong>\u00a0\u0394x = x <sub>final<\/sub>\u00a0 -x <sub>initial<\/sub> = x<sub>f<\/sub> - x<sub>o<\/sub><\/strong><\/p>\n<p style=\"text-align: left;padding-left: 30px\">where <strong><em>x<\/em><sub>0<\/sub><\/strong> is the initial position and <em><strong>x<sub>f<\/sub><\/strong><\/em> is the final position. In this text, the Greek letter\u00a0<strong>\u0394<\/strong> (delta) always means \"change in\" what ever quantity follows it. The SI unit for displacement is the meter (m). Displacement has a direction as well as a magnitude.<\/p>\n\n<ul>\n \t<li style=\"text-align: left\">When you start a problem, assign which direction will be positive.<\/li>\n \t<li style=\"text-align: left\">Distance is the magnitude of displacement between two positions.<\/li>\n \t<li style=\"text-align: left\">Distance traveled is the total length of the path traveled between two positions.<\/li>\n<\/ul>\n<\/div>\n<\/section>\n<div class=\"bcc-box bcc-info\">\n<h3>Conceptual Questions<\/h3>\n<section class=\"conceptual-questions\">\n<div id=\"fs-id1704056\" class=\"exercise\">\n<div id=\"fs-id1942381\" class=\"problem\">\n<p id=\"import-auto-id2774216\"><strong>1: <\/strong>Give an example in which there are clear distinctions among distance traveled, displacement, and magnitude of displacement. Specifically identify each quantity in your example.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id3147584\" class=\"exercise\">\n<div id=\"fs-id1702540\" class=\"problem\">\n<p id=\"import-auto-id1414240\"><strong>2: <\/strong>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\n<\/div>\n<\/div>\n<div id=\"fs-id3563423\" class=\"exercise\">\n<div id=\"fs-id2781126\" class=\"problem\">\n<p id=\"import-auto-id2173366\"><strong>3: <\/strong>Bacteria move back and forth by using their flagella (structures that look like little tails). Speeds of up to 50 \u03bcm\/s (50 \u00d7 10<sup>-6<\/sup> m\/s) have been observed. The total distance traveled by a bacterium is large for its size, while its displacement is small. Why is this?<\/p>\n\n<\/div>\n<\/div>\n<\/section><\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Problems &amp; Exercises<\/h3>\n&nbsp;\n<figure id=\"import-auto-id2076702\"><\/figure>\n<div id=\"fs-id1126076\" class=\"exercise\">\n<div id=\"fs-id3312138\" class=\"problem\">\n\n[caption id=\"attachment_4399\" align=\"aligncenter\" width=\"336\"]<a href=\"\/douglasphys1104summer2021\/chapter\/2-1-displacement\/figure_02_01sol_01-1-2\/\" rel=\"attachment wp-att-4399\"><img class=\"wp-image-75 size-full\" src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_02_01Sol_01-1-e1535573444727.jpg\" alt=\"Graph showing four paths.\" width=\"336\" height=\"207\"><\/a> Four objects travelling along a one dimensional path, with the distance axis labelled.[\/caption]\n<p id=\"import-auto-id3748212\"><strong>1: <\/strong>Find the following for path A: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id2823990\" class=\"exercise\">\n<div id=\"fs-id2804044\" class=\"problem\">\n<p id=\"import-auto-id1703980\"><strong>2: <\/strong>Find the following for path B: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id4292134\" class=\"exercise\">\n<div id=\"fs-id2811336\" class=\"problem\">\n<p id=\"import-auto-id4358788\"><strong>3: <\/strong>Find the following for path C: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id3242594\" class=\"exercise\">\n<div id=\"fs-id1470371\" class=\"problem\">\n<p id=\"import-auto-id2811759\"><strong>4: <\/strong>Find the following for path D: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<section>\n<div class=\"exercise\">\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1580711\" class=\"definition\">\n \t<dt>kinematics<\/dt>\n \t<dd id=\"fs-id1426394\">the study of motion without considering its causes<\/dd>\n<\/dl>\n<dl id=\"fs-id3059259\" class=\"definition\">\n \t<dt>position<\/dt>\n \t<dd id=\"fs-id1415754\">the location of an object at a particular time<\/dd>\n<\/dl>\n<dl id=\"fs-id2979435\" class=\"definition\">\n \t<dt>displacement<\/dt>\n \t<dd id=\"fs-id3222890\">the change in position of an object<\/dd>\n<\/dl>\n<dl id=\"fs-id3151040\" class=\"definition\">\n \t<dt>distance<\/dt>\n \t<dd id=\"fs-id3215864\">the magnitude of displacement between two positions<\/dd>\n<\/dl>\n<dl id=\"fs-id1548772\" class=\"definition\">\n \t<dt>distance traveled<\/dt>\n \t<dd id=\"fs-id1411294\">the total length of the path traveled between two positions<\/dd>\n<\/dl>\n<div class=\"bcc-box bcc-info\">\n<h3>Solutions<\/h3>\n<strong>Check Your Understanding 1<\/strong><span id=\"import-auto-id2928776\"><\/span>\n\n<section>\n<div class=\"exercise\">\n<div id=\"fs-id1711333\" class=\"solution\">\n<figure id=\"import-auto-id2776946\">\n\n[caption id=\"\" align=\"aligncenter\" width=\"350\"]<img src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_02_01_03-1.jpg\" alt=\"Two diagrams side by side. To the left is a horizontal line, or x axis, with points for final position and initial position. Displacement 1, shown by an arrow pointing leftward, equals negative 3 kilometers. Displacement 2, shown by an arrow pointing rightward, equals 2 kilometers. To the right is a pair of x and y axes, showing that east is the positive x direction and west is the negative x direction.\" width=\"350\" height=\"252\"> <strong>Figure 5.<\/strong>[\/caption]<\/figure>\n<p id=\"import-auto-id2892639\"><strong>1:<\/strong> (a) The rider\u2019s displacement is<strong>\u00a0\u0394x = x <sub>final<\/sub>\u00a0 -x <sub>initial<\/sub> = x<sub>f<\/sub> - x<sub>o<\/sub><\/strong>. The displacement is negative because we take east to be positive and west to be negative. \u00a0Or you could just say \"1 km to the West\". \u00a0Note that the drawing clearly showed that West was chosen to be negative. \u00a0 \u00a0(b) The distance traveled is 3 km + 2 km = 5 km. \u00a0 (c) The magnitude of the displacement is 1 km.<\/p>\n<strong>Problems &amp; Exercises<\/strong>\n\n<strong>1:\u00a0<\/strong>(a) 7 m \u00a0(b) 7 m (c) + 7 m\n\n<strong>2:<\/strong>\u00a0 (a) 5 m (b) 5 m (c) - 5 m\n\n<strong>3:\u00a0<\/strong>This is badly drawn so the answers are debatable. \u00a0Assuming it went from a position of 2 m to 10 then back to 8 and then back again to 10 m that gives a) distance of 12 m b) magnitude of the displacement as 8 m and c) a displacement of +8 m or 8 metres ot the right.\n\n<strong>4:<\/strong> \u00a0(a) \u00a08 m (b) 4 m (c) - 4 m\n\n<\/div>\n<\/div>\n<\/section><\/div>\n<\/div>\n<\/section>","rendered":"<figure id=\"import-auto-id2723149\"><figcaption><\/figcaption><div class=\"bcc-box bcc-highlight\">\n<h3>Summary<\/h3>\n<ul>\n<li>Define position, displacement, distance, and distance traveled.<\/li>\n<li>Explain the relationship between position and displacement.<\/li>\n<li>Distinguish between displacement and distance traveled.<\/li>\n<li>Calculate displacement and distance given initial position, final position and the path between the two.<\/li>\n<\/ul>\n<\/div>\n<figure style=\"width: 365px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104\/wp-content\/uploads\/sites\/1393\/2017\/06\/Figure_02_01_00-1.jpg\" alt=\"Three people cycling along a canal. The blurred buildings in the background convey a sense of motion of the cyclists.\" width=\"365\" height=\"274\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 1.<\/strong> These cyclists in Vietnam can be described by their position relative to buildings and a canal. Their motion can be described by their change in position, or displacement, in the frame of reference. (credit: Suzan Black, Fotopedia).<\/figcaption><\/figure>\n<\/figure>\n<section id=\"fs-id3178358\">\n<h1>Position<\/h1>\n<p id=\"import-auto-id1572700\">In order to describe the motion of an object, you must first be able to describe its <strong>position<\/strong>\u2014where it is at any particular time. More precisely, you need to specify its position relative to a convenient reference frame. Earth is often used as a reference frame, and we often describe the position of an object as it relates to stationary objects in that reference frame. For example, a rocket launch would be described in terms of the position of the rocket with respect to the Earth as a whole, while a professor\u2019s position could be described in terms of where she is in relation to the nearby white board. \u00a0In other cases, we use reference frames that are not stationary but are in motion relative to the Earth. To describe the position of a person in an airplane, for example, we use the airplane, not the Earth, as the reference frame. Please see the figures below.<\/p>\n<\/section>\n<section id=\"fs-id2572460\">\n<h1>Displacement<\/h1>\n<p id=\"import-auto-id3310684\">If an object moves relative to a reference frame (for example, if a professor moves to the right relative to a white board or a passenger moves toward the rear of an airplane), then the object\u2019s position changes. This change in position is known as <strong>displacement<\/strong>. The word \u201cdisplacement\u201d implies that an object has moved, or has been displaced.<\/p>\n<div id=\"fs-id3206000\" class=\"note\">\n<div class=\"textbox shaded\">\n<h3>DISPLACEMENT<\/h3>\n<p id=\"import-auto-id2986917\">Displacement is the <em>change in position<\/em> of an object:<\/p>\n<div id=\"eip-458\" class=\"equation\" style=\"text-align: center\"><strong>\u0394<em>x =x<sub>f\u00a0\u00a0<\/sub>\u00a0&#8211;\u00a0x<sub>0<\/sub><\/em><\/strong><\/div>\n<div class=\"equation\" style=\"text-align: center\"><\/div>\n<p id=\"import-auto-id2795085\">where\u00a0<strong>\u0394<em>x<\/em><\/strong> is displacement, <em><strong>x<sub>f<\/sub><\/strong><\/em> is the final position, and <strong><em>x<\/em><sub>0<\/sub><\/strong> is the initial position.<\/p>\n<\/div>\n<p id=\"import-auto-id2804940\">In this text the upper case Greek letter\u00a0<strong>\u0394<\/strong> (delta) always means \u201cchange in\u201d whatever quantity follows it; thus, <strong>\u0394<em>x<\/em><\/strong> means <em>change in position<\/em>. Always solve for displacement by subtracting initial position <strong><em>x<\/em><sub>0<\/sub><\/strong> from final <em><strong>x<sub>f<\/sub><\/strong><\/em>.<em><br \/>\n<\/em><\/p>\n<p id=\"import-auto-id2880746\">Note that the SI unit for displacement is the meter (m) (see <a href=\"\/douglasphys1104summer2021\/part\/chapter-1-the-nature-of-science-and-physics\/\">Chapter 1 Physical Quantities and Units<\/a>), but sometimes kilometers, miles, feet, and other units of length are used. Keep in mind that when units other than the meter are used in a problem, you may need to convert them into meters to complete the calculation.<\/p>\n<\/div>\n<figure id=\"attachment_4397\" aria-describedby=\"caption-attachment-4397\" style=\"width: 705px\" class=\"wp-caption aligncenter\"><a href=\"\/douglasphys1104summer2021\/chapter\/2-1-displacement\/figure_02_01_01-1-2\/\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-73 size-full\" src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_02_01_01-1.jpg\" alt=\"Figure 2. Professor walking in front of a chalkboard.\" width=\"705\" height=\"560\" srcset=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_02_01_01-1.jpg 705w, https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_02_01_01-1-300x238.jpg 300w, https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_02_01_01-1-65x52.jpg 65w, https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_02_01_01-1-225x179.jpg 225w, https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_02_01_01-1-350x278.jpg 350w\" sizes=\"auto, (max-width: 705px) 100vw, 705px\" \/><\/a><figcaption id=\"caption-attachment-4397\" class=\"wp-caption-text\"><a id=\"\u201dFigure\" href=\"\"><\/a>Figure 2. 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.<\/figcaption><\/figure>\n<\/section>\n<p>&nbsp;<\/p>\n<section id=\"fs-id2572460\">\n<div id=\"fs-id3206000\" class=\"note\">\n<figure id=\"import-auto-id2707699\">\n<figure style=\"width: 541px\" class=\"wp-caption aligncenter\"><a href=\"\/wp-content\/uploads\/sites\/153\/2017\/06\/Figure_02_01_02-1.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_02_01_02-1.jpg\" alt=\"View of an airplane with an inset of the passengers sitting inside. A passenger has just moved from his seat and is now standing in the back. His initial position was 6 point 0 meters. His final position is 2 point 0 meters. His displacement is given by the equation delta x equals x sub f minus x sub 0 equals 4 point zero meters.\" width=\"541\" height=\"399\" \/><\/a><figcaption class=\"wp-caption-text\"><strong>Figure 3.<\/strong> A passenger moves from his seat to the back of the plane. His location relative to the airplane is given by <em><strong>x<\/strong><\/em>. The <strong>-4.0-m<\/strong> displacement of the passenger relative to the plane is represented by an arrow toward the rear of the plane. Notice that the arrow representing his displacement is twice as long as the arrow representing the displacement of the professor shown above (he moves twice as far).<\/figcaption><\/figure>\n<\/figure>\n<p id=\"import-auto-id2815802\">Note that displacement has a direction as well as a magnitude. The professor\u2019s displacement is 2.0 m to the right, and the airline passenger\u2019s displacement is 4.0 m toward the rear. In one-dimensional motion, direction can be specified with a plus or minus sign. When you begin a problem, you should select which direction is positive (usually that will be to the right or up, but you are free to select positive as being any direction). The professor\u2019s initial position is <strong><em>x<\/em><sub>0<\/sub> = 1.5 m<\/strong> and her final position is <strong><em>x<sub>f<\/sub><\/em> = 3.5 m<\/strong>. Thus her displacement is<\/p>\n<div id=\"eip-556\" class=\"equation\" style=\"text-align: center\"><strong>\u00a0\u0394x = x <sub>final<\/sub>\u00a0 -x <sub>initial<\/sub> = x<sub>f<\/sub> &#8211; x<sub>o<\/sub><\/strong> = 3.5 m -1.5 m\u00a0 = +2.0 m<\/div>\n<p id=\"import-auto-id1373168\">In this coordinate system, motion to the right is positive, whereas motion to the left is negative. Similarly, the airplane passenger\u2019s initial position is <strong><em>x<\/em><sub>0<\/sub> = 6.0 m<\/strong> and his final position is <strong><em>x<sub>f<\/sub><\/em> = 2.0 m<\/strong>, so his displacement is<\/p>\n<\/div>\n<p style=\"text-align: center\"><strong>\u00a0\u0394x = x <sub>final<\/sub>\u00a0 -x <sub>initial<\/sub> = x<sub>f<\/sub> &#8211; x<sub>o<\/sub><\/strong> = 2.0 m &#8211; 6.0\u00a0 m\u00a0 = -4.0 m<\/p>\n<div id=\"fs-id3206000\" class=\"note\">\n<p id=\"import-auto-id3579333\">His displacement is negative because his motion is toward the rear of the plane, or in the negative x direction in our coordinate system.<\/p>\n<\/div>\n<\/section>\n<section id=\"fs-id1414454\">\n<h1>Distance<\/h1>\n<p id=\"import-auto-id3580245\">Although displacement is described in terms of direction, distance is not. <strong>Distance <\/strong>is defined to be <em>the magnitude or size of displacement between two positions<\/em>. Note that the distance between two positions is not the same as the distance traveled between them. <strong>Distance traveled<\/strong> is <em>the total length of the path traveled between two positions<\/em>. Distance has no direction and, thus, no sign. For example, the distance the professor walks is 2.0 m. The distance the airplane passenger walks is 4.0 m.<\/p>\n<div id=\"fs-id3044492\" class=\"note\">\n<div class=\"textbox shaded\">\n<h3>MISCONCEPTION ALERT: DISTANCE TRAVELED VS. MAGNITUDE OF DISPLACEMENT<\/h3>\n<p id=\"import-auto-id2850966\">It is important to note that the <em>distance traveled<\/em>, however, can be greater than the magnitude of the displacement (by magnitude, we mean just the size of the displacement without regard to its direction; that is, just a number with a unit). For example, the professor could pace back and forth many times, perhaps walking a distance of 150 m during a lecture, yet still end up only 2.0 m to the right of her starting point. In this case her displacement would be +2.0 m, the magnitude of her displacement would be 2.0 m, but the distance she traveled would be 150 m. In kinematics we nearly always deal with displacement and magnitude of displacement, and almost never with distance traveled. One way to think about this is to assume you marked the start of the motion and the end of the motion. The displacement is simply the difference in the position of the two marks and is independent of the path taken in traveling between the two marks. The distance traveled, however, is the total length of the path taken between the two marks.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id3589986\" class=\"exercise\">\n<div class=\"bcc-box bcc-info\">\n<h3>Check Your Understanding 1<\/h3>\n<div id=\"fs-id2996632\" class=\"problem\">\n<p id=\"import-auto-id3225677\"><strong>1:<\/strong> A cyclist rides 3 km west and then turns around and rides 2 km east. (a) What is her displacement? (b) What distance does she ride? (c) What is the magnitude of her displacement?<\/p>\n<\/div>\n<\/div>\n<h1>\u00a0Section Summary<\/h1>\n<ul>\n<li>Kinematics is the study 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.<\/li>\n<li>In symbols, displacement\u00a0<strong>\u0394<em>x<\/em><\/strong> is defined to be<\/li>\n<\/ul>\n<p style=\"text-align: center\"><strong>\u00a0\u0394x = x <sub>final<\/sub>\u00a0 -x <sub>initial<\/sub> = x<sub>f<\/sub> &#8211; x<sub>o<\/sub><\/strong><\/p>\n<p style=\"text-align: left;padding-left: 30px\">where <strong><em>x<\/em><sub>0<\/sub><\/strong> is the initial position and <em><strong>x<sub>f<\/sub><\/strong><\/em> is the final position. In this text, the Greek letter\u00a0<strong>\u0394<\/strong> (delta) always means &#8220;change in&#8221; what ever quantity follows it. The SI unit for displacement is the meter (m). Displacement has a direction as well as a magnitude.<\/p>\n<ul>\n<li style=\"text-align: left\">When you start a problem, assign which direction will be positive.<\/li>\n<li style=\"text-align: left\">Distance is the magnitude of displacement between two positions.<\/li>\n<li style=\"text-align: left\">Distance traveled is the total length of the path traveled between two positions.<\/li>\n<\/ul>\n<\/div>\n<\/section>\n<div class=\"bcc-box bcc-info\">\n<h3>Conceptual Questions<\/h3>\n<section class=\"conceptual-questions\">\n<div id=\"fs-id1704056\" class=\"exercise\">\n<div id=\"fs-id1942381\" class=\"problem\">\n<p id=\"import-auto-id2774216\"><strong>1: <\/strong>Give an example in which there are clear distinctions among distance traveled, displacement, and magnitude of displacement. Specifically identify each quantity in your example.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id3147584\" class=\"exercise\">\n<div id=\"fs-id1702540\" class=\"problem\">\n<p id=\"import-auto-id1414240\"><strong>2: <\/strong>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-id3563423\" class=\"exercise\">\n<div id=\"fs-id2781126\" class=\"problem\">\n<p id=\"import-auto-id2173366\"><strong>3: <\/strong>Bacteria move back and forth by using their flagella (structures that look like little tails). Speeds of up to 50 \u03bcm\/s (50 \u00d7 10<sup>-6<\/sup> m\/s) have been observed. The total distance traveled by a bacterium is large for its size, while its displacement is small. Why is this?<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Problems &amp; Exercises<\/h3>\n<p>&nbsp;<\/p>\n<figure id=\"import-auto-id2076702\"><\/figure>\n<div id=\"fs-id1126076\" class=\"exercise\">\n<div id=\"fs-id3312138\" class=\"problem\">\n<figure id=\"attachment_4399\" aria-describedby=\"caption-attachment-4399\" style=\"width: 336px\" class=\"wp-caption aligncenter\"><a href=\"\/douglasphys1104summer2021\/chapter\/2-1-displacement\/figure_02_01sol_01-1-2\/\" rel=\"attachment wp-att-4399\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-75 size-full\" src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_02_01Sol_01-1-e1535573444727.jpg\" alt=\"Graph showing four paths.\" width=\"336\" height=\"207\" srcset=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_02_01Sol_01-1-e1535573444727.jpg 336w, https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_02_01Sol_01-1-e1535573444727-300x185.jpg 300w, https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_02_01Sol_01-1-e1535573444727-65x40.jpg 65w, https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_02_01Sol_01-1-e1535573444727-225x139.jpg 225w\" sizes=\"auto, (max-width: 336px) 100vw, 336px\" \/><\/a><figcaption id=\"caption-attachment-4399\" class=\"wp-caption-text\">Four objects travelling along a one dimensional path, with the distance axis labelled.<\/figcaption><\/figure>\n<p id=\"import-auto-id3748212\"><strong>1: <\/strong>Find the following for path A: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2823990\" class=\"exercise\">\n<div id=\"fs-id2804044\" class=\"problem\">\n<p id=\"import-auto-id1703980\"><strong>2: <\/strong>Find the following for path B: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id4292134\" class=\"exercise\">\n<div id=\"fs-id2811336\" class=\"problem\">\n<p id=\"import-auto-id4358788\"><strong>3: <\/strong>Find the following for path C: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id3242594\" class=\"exercise\">\n<div id=\"fs-id1470371\" class=\"problem\">\n<p id=\"import-auto-id2811759\"><strong>4: <\/strong>Find the following for path D: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<section>\n<div class=\"exercise\">\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1580711\" class=\"definition\">\n<dt>kinematics<\/dt>\n<dd id=\"fs-id1426394\">the study of motion without considering its causes<\/dd>\n<\/dl>\n<dl id=\"fs-id3059259\" class=\"definition\">\n<dt>position<\/dt>\n<dd id=\"fs-id1415754\">the location of an object at a particular time<\/dd>\n<\/dl>\n<dl id=\"fs-id2979435\" class=\"definition\">\n<dt>displacement<\/dt>\n<dd id=\"fs-id3222890\">the change in position of an object<\/dd>\n<\/dl>\n<dl id=\"fs-id3151040\" class=\"definition\">\n<dt>distance<\/dt>\n<dd id=\"fs-id3215864\">the magnitude of displacement between two positions<\/dd>\n<\/dl>\n<dl id=\"fs-id1548772\" class=\"definition\">\n<dt>distance traveled<\/dt>\n<dd id=\"fs-id1411294\">the total length of the path traveled between two positions<\/dd>\n<\/dl>\n<div class=\"bcc-box bcc-info\">\n<h3>Solutions<\/h3>\n<p><strong>Check Your Understanding 1<\/strong><span id=\"import-auto-id2928776\"><\/span><\/p>\n<section>\n<div class=\"exercise\">\n<div id=\"fs-id1711333\" class=\"solution\">\n<figure id=\"import-auto-id2776946\">\n<figure style=\"width: 350px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_02_01_03-1.jpg\" alt=\"Two diagrams side by side. To the left is a horizontal line, or x axis, with points for final position and initial position. Displacement 1, shown by an arrow pointing leftward, equals negative 3 kilometers. Displacement 2, shown by an arrow pointing rightward, equals 2 kilometers. To the right is a pair of x and y axes, showing that east is the positive x direction and west is the negative x direction.\" width=\"350\" height=\"252\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 5.<\/strong><\/figcaption><\/figure>\n<\/figure>\n<p id=\"import-auto-id2892639\"><strong>1:<\/strong> (a) The rider\u2019s displacement is<strong>\u00a0\u0394x = x <sub>final<\/sub>\u00a0 -x <sub>initial<\/sub> = x<sub>f<\/sub> &#8211; x<sub>o<\/sub><\/strong>. The displacement is negative because we take east to be positive and west to be negative. \u00a0Or you could just say &#8220;1 km to the West&#8221;. \u00a0Note that the drawing clearly showed that West was chosen to be negative. \u00a0 \u00a0(b) The distance traveled is 3 km + 2 km = 5 km. \u00a0 (c) The magnitude of the displacement is 1 km.<\/p>\n<p><strong>Problems &amp; Exercises<\/strong><\/p>\n<p><strong>1:\u00a0<\/strong>(a) 7 m \u00a0(b) 7 m (c) + 7 m<\/p>\n<p><strong>2:<\/strong>\u00a0 (a) 5 m (b) 5 m (c) &#8211; 5 m<\/p>\n<p><strong>3:\u00a0<\/strong>This is badly drawn so the answers are debatable. \u00a0Assuming it went from a position of 2 m to 10 then back to 8 and then back again to 10 m that gives a) distance of 12 m b) magnitude of the displacement as 8 m and c) a displacement of +8 m or 8 metres ot the right.<\/p>\n<p><strong>4:<\/strong> \u00a0(a) \u00a08 m (b) 4 m (c) &#8211; 4 m<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":9,"menu_order":1,"comment_status":"closed","ping_status":"closed","template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-77","chapter","type-chapter","status-publish","hentry"],"part":71,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/pressbooks\/v2\/chapters\/77","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/wp\/v2\/users\/9"}],"replies":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/wp\/v2\/comments?post=77"}],"version-history":[{"count":1,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/pressbooks\/v2\/chapters\/77\/revisions"}],"predecessor-version":[{"id":78,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/pressbooks\/v2\/chapters\/77\/revisions\/78"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/pressbooks\/v2\/parts\/71"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/pressbooks\/v2\/chapters\/77\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/wp\/v2\/media?parent=77"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/pressbooks\/v2\/chapter-type?post=77"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/wp\/v2\/contributor?post=77"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/wp\/v2\/license?post=77"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}