22. DisplacementLearning Objectives
PositionIn order to describe the motion of an object, you must first be able to describe its DisplacementIf 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’s position changes. This change in position is known as DisplacementDisplacement is the change in position of an object: $$\text{\Delta}x={x}_{\text{f}}{x}_{0},$$where $\text{\Delta}x$ is displacement, ${x}_{\text{f}}$ is the final position, and ${x}_{0}$ is the initial position. In this text the upper case Greek letter $\text{\Delta}$ (delta) always means “change in” whatever quantity follows it; thus, $\text{\Delta}x$ means change in position. 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 (m) (see Physical Quantities and Units), 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. Note that displacement has a direction as well as a magnitude. The professor’s displacement is 2.0 m to the right, and the airline passenger’s displacement is 4.0 m toward the rear. In onedimensional 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’s initial position is ${x}_{0}=1\text{.}5\phantom{\rule{0.25em}{0ex}}\text{m}$ and her final position is ${x}_{\text{f}}=3\text{.}5\phantom{\rule{0.25em}{0ex}}\text{m}$. Thus her displacement is $$\mathrm{\Delta}x={x}_{\mathrm{f}}{x}_{0}=3\text{.5 m}\mathrm{1.5\; m}=+2\text{.0 m}.$$In this coordinate system, motion to the right is positive, whereas motion to the left is negative. Similarly, the airplane passenger’s initial position is ${x}_{0}=6\text{.}\mathrm{0\; m}$ and his final position is ${x}_{\mathrm{f}}=2\text{.}\mathrm{0\; m}$, so his displacement is $$\mathrm{\Delta}x={x}_{\mathrm{f}}{x}_{0}=2\text{.}\mathrm{0\; m}6\text{.}\mathrm{0\; m}=4\text{.}\mathrm{0\; m}.$$His displacement is negative because his motion is toward the rear of the plane, or in the negative $x$ direction in our coordinate system. DistanceAlthough displacement is described in terms of direction, distance is not. Misconception Alert: Distance Traveled vs. Magnitude of DisplacementIt is important to note that the distance traveled, 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. Check Your UnderstandingA 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? Show/Hide Solution SolutionFigure 4 (a) The rider’s displacement is $\mathrm{\Delta}x={x}_{\mathrm{f}}{x}_{0}=\text{\u22121 km}$. (The displacement is negative because we take east to be positive and west to be negative.) (b) The distance traveled is $\text{3 km}+\text{2 km}=\text{5 km}$. (c) The magnitude of the displacement is $\mathrm{1\; km}$. Section Summary
Conceptual QuestionsExercise 1Give an example in which there are clear distinctions among distance traveled, displacement, and magnitude of displacement. Specifically identify each quantity in your example. Exercise 2Under 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? Exercise 3Bacteria move back and forth by using their flagella (structures that look like little tails). Speeds of up to $\text{50 \mu m/s}\phantom{\rule{0.25em}{0ex}}\left(\text{50}\times {\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)$ have been observed. The total distance traveled by a bacterium is large for its size, while its displacement is small. Why is this? Problems & ExercisesFigure 5 Exercise 1Find the following for path A in Figure 5: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish. Show/Hide Solution Solution(a) 7 m (b) 7 m (c) $+\mathrm{7\; m}$ Exercise 2Find the following for path B in Figure 5: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish. Exercise 3Find the following for path C in Figure 5: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish. Show/Hide Solution Solution(a) 13 m (b) 9 m (c) $+\mathrm{9\; m}$ Exercise 4Find the following for path D in Figure 5: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.
