By the end of this lesson, you should be able to:

- Define position, displacement, distance, and distance traveled.
- 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.

## Position

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. We often use Earth 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’s position could be described in terms of where she is in relation to the nearby white board. (See image above.) In 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. (See image below.)

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 \(x_0 = 1.5 \text{ m}\) and her final position is \(x_{\text{f}} = 3.5 \text{ m}\). Thus her displacement is

\(\Delta x = x_{\text{f}} − x_0 = \)\(3.5\text{ m} − 1.5\text{ m} = +2.0\text{ m}.\)

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 \(x_{\text{f}} = 6.0 \text{ m}\) and his final position is \(x_{\text{f}} = 2.0 \text{ m}\), so his displacement is

\(\Delta x = x_{\text{f}} − x_0 = \)\(2.0\text{ m} − 6.0\text{ m} = −4.0\text{ m}.\)

His displacement is negative because his motion is toward the rear of the plane, or in the negative *x* direction in our coordinate system.

## Distance

Although displacement is described in terms of direction, distance is not. **Distance** is defined to be *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, 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.