# Distance and Displacement in Physics: Definition and Examples

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• 0:01 5K: How Far Did I Travel?
• 0:35 Distance vs. Displacement
• 2:22 Calculating Displacement
• 4:24 Lesson Summary

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Lesson Transcript
Instructor: Angela Hartsock

Angela has taught college Microbiology and has a doctoral degree in Microbiology.

Distance and displacement might seem like similar terms but in physics, understanding the difference can mean getting a question right instead of wrong. In this lesson, we will define these terms and illustrate how easy it is to confuse the two.

## 5K: How Far Did I Travel?

Last Thanksgiving, I decided to run my first 5K. I won't embarrass myself by telling you my exact time, but I was able to successfully finish the race. If you asked the other racers how far I traveled, they would all tell you 5 kilometers. If you asked a group of physicists how far I traveled, they might all tell you 0 kilometers - that I didn't change position at all. It might surprise you to learn that both answers are correct. But how can that be? It all depends on the subtle yet very important difference between distance and displacement.

## Distance Versus Displacement

Let's start by defining these two terms, starting with the easier of the two.

Distance is how much ground is covered by an object, regardless of its starting or ending position. There is no directional component to a distance measurement, making it a scalar quantity. So, during my 5K, I ran 5 kilometers total. It doesn't matter where the starting line or finish line were or in which direction I was running. It only matters that, if you trace and measure my path, I covered a distance of 5 km of ground. So, when we asked the other runners how far I had traveled, they all answered correctly with a distance measurement of 5 kilometers.

Let's take a minute to look at our other term: displacement, which is an object's change in position considering its starting position and final position. A displacement measurement does not take into account what route the object took to change position, only where it started and where it ended. It is easiest to picture displacement by locating where the object started and drawing a straight arrow from this point to the point where the object stopped moving. Remember, in physics this arrow is called a vector. Its length corresponds to the magnitude, or size, of the movement, and the arrow points in the direction of travel. This makes displacement a vector quantity because it incorporates both movement magnitude and direction.

In the shorthand of physics, displacement is written as Î”s. 'Delta' is a Greek letter shaped like a triangle and it's used to represent 'change in.' The 's' stands for spatial location. So Î”s stands for a change in spatial location. You should get comfortable using this shorthand just in case you see questions asking for you to solve for Î”s that don't actually ask for displacement by name.

## Calculating Displacement

Now, before we look at my displacement during my 5K, I want to try and illustrate the concept of displacement. If I walk 3 meters west, then turn and walk 4 meters north, you can easily calculate the distance traveled: 3 m + 4 m = 7 meters. Since this is a distance, there is no need to worry about the direction of travel. Now, let's determine my displacement. I can draw an arrow that starts where I started walking and extends to the point where I stopped walking. It should look like this, forming a right triangle. To determine my displacement, all I need to do is determine how long the arrow is using the lengths of the sides and the Pythagorean Theorem.

a^2 + b^2 = c^2

Since c^2 is your displacement squared, it should be written as (Î”s)^2.

Î”s = âˆš{(3*3) + (4*4)}

Î”s = âˆš(9 + 16)

Î”s = 5 meters northwest

Doing the math yields a value of 5 meters. But this is not the displacement. Remember, we need a direction as well. In this case, I ended up northwest of where I started, so my displacement is 5 meters northwest. Notice that I traveled a total distance of 7 m but have only a displacement of 5 meters northwest.

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