Learning Objectives

Learning Objectives

By the end of this section, you will be able to do the following:

  • Define position, displacement, distance, and distance traveled in a particular frame of reference
  • Explain the relationship between position and displacement
  • Distinguish between displacement and distance traveled
  • Calculate displacement and distance given initial position, final position, and the path between the two

The information presented in this section supports the following AP® learning objectives and science practices:

  • 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2)
  • 3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1)
Three people cycling along a canal. The blurred buildings in the background convey a sense of motion of the cyclists.
Figure 2.2 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)



In order to describe the motion of an object, you must first be able to describe its position—where 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 Earth as a whole, while a professor's position could be described in terms of where she is in relation to the nearby whiteboard (see Figure 2.3). 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 (see Figure 2.4).



If an object moves relative to a reference frame, for example, if a professor moves to the right relative to a whiteboard or a passenger moves toward the rear of an airplane, then the object's position changes. This change in position is known as displacement. The word displacement implies that an object has moved, or has been displaced.


Displacement is the change in position of an object

2.1 Δx=xfx0,Δx=xfx0, size 12{Δx=x rSub { size 8{f} } - x rSub { size 8{0} } } {}

where ΔxΔx size 12{Δx} {} is displacement, xfxf size 12{x rSub { size 8{f} } } {} is the final position, and x0x0 size 12{x rSub { size 8{0} } } {} is the initial position.

In this text, the uppercase Greek letter ΔΔ size 12{Δ} {} (delta) always means change in whatever quantity follows it; thus, ΔxΔx size 12{Δx} {} means change in position. Always solve for displacement by subtracting initial position x0x0 size 12{x rSub { size 8{0} } } {} from final position xf.xf. size 12{x rSub { size 8{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.

The initial and final position of a professor as she moves to the right while writing on a whiteboard. Her initial position is 1 point 5 meters. Her final position is 3 point 5 meters. Her displacement is given by the equation delta x equals x sub f minus x sub 0 equals 2 point 0 meters.
Figure 2.3 A professor paces left and right while lecturing. Her position relative to the blackboard is given by xx size 12{x} {}. The +2.0-m+2.0-m size 12{+2 "." 0`m} {} displacement of the professor relative to the blackboard is represented by an arrow pointing to the right.
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.
Figure 2.4 A passenger moves from his seat to the back of the plane. His location relative to the airplane is given by xx size 12{x} {}. The −4-m 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—he moves twice as far—in Figure 2.3.

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 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's initial position is x0=1.5mx0=1.5m size 12{x rSub { size 8{0} } =1 "." 5`m} {} and her final position is xf=3.5mxf=3.5m size 12{x rSub { size 8{f} } =3 "." 5`m} {}. Thus her displacement is

2.2 Δx=xfx0=3.5 m1.5 m =+2.0 m.Δx=xfx0=3.5 m1.5 m =+2.0 m. size 12{Δx=x"" lSub { size 8{f} } - x rSub { size 8{0} } =3 "." 5`m - 1 "." 5`"m "= +2 "." "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 x0=6.0 mx0=6.0 m and his final position is xf=2.0 mxf=2.0 m size 12{x rSub { size 8{f} } =2 "." 0`m} {}, so his displacement is

2.3 Δx=xfx0=2.0 m6.0 m=4.0 m.Δx=xfx0=2.0 m6.0 m=4.0 m. size 12{Δx=x"" lSub { size 8{f} } - x rSub { size 8{0} } =2 "." 0`m - 6 "." 0`m= - 4 "." 0`m} {}

His displacement is negative because his motion is toward the rear of the plane, or in the negative x-x- size 12{x} {}direction in our coordinate system.



Although displacement is described in terms of direction, distance is not. Distance is defined as the magnitude or size of displacement between two positions. Note that the distance between two positions is not the same as the distance traveled between them. Distance traveled is the total length of the path traveled between two positions. 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.

Misconception Alert: Distance Traveled vs. Magnitude of Displacement

It 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 and 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 Understanding

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?


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.
Figure 2.5

(a) The rider's displacement is Δx=xfx0=−1 kmΔx=xfx0=−1 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 km3 km + 2 km = 5 km size 12{"3 km "+" 2 km "=" 5 km"} {}.

(c) The magnitude of the displacement is 1 km1 km size 12{1" km"} {}.