Analyzing motion can get complicated. Learning precise vocabulary will help.

What does position mean?

In physics, we love to precisely describe the motion of an object. Seriously, the first few chapters of basically every physics textbook are devoted to teaching people how to precisely describe motion since it is so important to everything else we do in physics.
But to describe an object's motion, we have to first be able to describe its position—where it is at any particular time. More precisely, we 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 professor’s position could be described in terms of where she is in relation to the nearby white board (Figure 1). In other cases, we use reference frames that are not stationary but rather 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 (Figure 2).
The variable xx is often used to represent the horizontal position. The variable yy is often used to represent the vertical position.
The variable zz is used to describe the third perpendicular axis that typically points "out of the screen/page". But trying to draw motion that is coming out of the plane of a piece of paper or the screen is hard and a little awkward looking. So we typically don't deal with the zz direction unless it is particularly important to do so.
Besides, all the tools used to describe motion in the xx and yy directions work equally well for the zz direction as well, so we aren't leaving much out by neglecting the zz direction.

What does displacement mean?

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 defined to be the change in position of an object. It can be defined mathematically with the following equation:
Displacement=Δx=xfx0\text{Displacement}=\Delta x=x_f-x_0
xfx_f refers to the value of the final position.
x0x_0 refers to the value of the initial position.
Δx\Delta x is the symbol used to represent displacement.
The triangle symbol Δ\Delta is the Greek capital letter delta. It is used in physics to represent the change in a variable. When placed before a variable (yy) it refers to the difference between the final value (yfy_f) and initial value (y0y_0) of that variable.
Δy=yfy0\Delta y=y_f-y_0
I have to warn you that many people erroneously subtract the final value from the initial value y0yfy_0-y_f when finding difference, which would give Δy-\Delta y. This is likely because the initial value is often given first in a problem so people naturally put it first in the equation. You have to make sure you subtract the initial value from the final value and not the other way around.
Also you should know that some people use the subscript "0" to refer to the initial value (e.g., x0x_0). Other people use the letter "i" (e.g., xix_i) or the number "1" (e.g., x1x_1) to refer to initial value. But they are all just different ways of saying the same thing.
Displacement is a vector. This means it has a direction as well as a magnitude and is represented visually as an arrow that points from the initial position to the final position. For example, consider the professor that walks relative to the whiteboard in Figure 1.

Figure 1: A professor paces left and right while lecturing. The +2.0 m+2.0\text{ m} displacement of the professor relative to the whiteboard is represented by an arrow pointing to the right. (Image credit: Openstax College Physics)
The professor’s initial position is x0=1.5 mx_0=1.5\text{ m} and her final position is xf=3.5 mx_f=3.5\text{ m}. Thus, her displacement can be found as follows, . In this coordinate system, motion to the right is positive, whereas motion to the left is negative.
Now consider the passenger that walks relative to the plane in Figure 2.
Figure 2: A passenger moves from his seat to the back of the plane. The displacement of the passenger relative to the plane is represented by an arrow toward the rear of the plane. (Image credit: Openstax College Physics)
The airplane passenger’s initial position is x0=6.0 mx_0=6.0\text{ m} and his final position is xf=2.0 mx_f=2.0\text{ m}, so his displacement can be found as follows, . His displacement is negative because his motion is toward the rear of the plane, or in the negative x direction in our coordinate system.
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.

What do distance and distance traveled mean?

We must be careful when using the word distance since there are two ways in which the term distance is used in physics. We can talk about the distance between two points, or we can talk about the distance traveled by an object.
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 traveled is not a vector. It has no direction and, thus, no negative sign. For example, the distance the professor walks is 2.0 m2.0 \text{ m}. The distance the airplane passenger walks is 4.0 m4.0 \text{ m}.
It is important to note that the distance traveled does not have to equal the magnitude of the displacement (i.e., distance between the two points). Specifically, if an object changes direction in its journey, the total distance traveled will be greater than the magnitude of the displacement between those two points. See the solved examples below.

What's confusing about displacement?

People often forget that the distance traveled can be greater than the magnitude of the displacement. By magnitude, we mean the size of the displacement without regard to its direction (i.e., just a number with a unit). For example, the professor could pace back and forth many times, perhaps walking a distance of 150 meters during a lecture, yet still end up only two meters to the right of her starting point. In this case her displacement would be +2 m+2 \text{ m}, the magnitude of her displacement would be 2 m2 \text{ m}, but the distance she traveled would be 150 m150 \text{ 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 when traveling between the two marks. The distance traveled, however, is the total length of the path taken between the two marks.
People often forget to include a negative sign, if needed, in their answer for displacement. This sometimes occurs if they accidentally subtract the final position from the initial position rather than subtracting the initial position from the final position.

What do solved examples involving displacement look like?

Example 1: Displacement of four moving objects

Four objects move according to the paths shown in the diagram below. Assume the units of the horizontal scale are given in meters. (Image credit: altered from Openstax College Physics)
What was the displacement of each object?
Object A had an initial position of 0 m0\text{ m} and a final position of 7 m7\text{ m}. The displacement of object A can be shown with this equation:
ΔxA=7 m0 m=+7 m\Delta x_A= 7\text{ m}-0\text{ m}=+7\text{ m}
Object B had an initial position of 12 m12\text{ m} and a final position of 7 m7\text{ m}. The displacement of object B can be shown with this equation:
ΔxB=7 m12 m=5 m\Delta x_B= 7\text{ m}-12\text{ m}=-5\text{ m}
Object C had an initial position of 2 m2\text{ m} and a final position of 10 m10\text{ m}. The displacement of object C can be shown with this equation:
ΔxC=10 m2 m=+8 m\Delta x_C= 10\text{ m}-2\text{ m}=+8\text{ m}
Object D had an initial position of 9 m9\text{ m} and a final position of 5 m5\text{ m}. The displacement of object D can be shown with this equation:
ΔxD=5 m9 m=4 m\Delta x_D= 5\text{ m}-9\text{ m}=-4\text{ m}

Example 2: Distance traveled of four moving objects

Four objects move according to the paths shown in the diagram below. Assume the units of the horizontal scale are given in meters. (Image credit: altered from Openstax College Physics)
What was the total distance traveled by each object?
Object A travels a total distance of 7 m7\text{ m}.
Object B travels a total distance of 5 m5\text{ m}.
Object C travels a total distance of 8 m+2 m+2 m=12 m8\text{ m}+2\text{ m}+2\text{ m}=12\text{ m}.
Object D travels a total distance of 6 m+2 m=8 m6\text{ m}+2\text{ m}=8\text{ m}.
This article was adapted from the following article:
  1. "Displacement" from Openstax College Physics. Download the original article free at http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a@9.1:8/Displacement
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