# How To Calculate Displacement

Displacement is a quantity that is very useful. In physics we want to precisely calculate the position of an object (where it is at any given time). In many cases we want to know the position relative to some reference. This change of the object’s position is called displacement, and it tells us how far an object is relative to its starting position.

Depending on what data we are given, we can calculate the displacement in different ways:

-Given the starting and final positions.

-Given the velocity and time.

-Given the velocity, time and acceleration.

This article will cover how to calculate displacement for all of those three cases, and it will also explain the difference between distance and displacement.

# What is displacement?

Displacement is often confused with distance, but they are not the same thing. Distance refers to how much ground an object has traveled. Displacement, on the other hand, tells us how far out of place an object is.

We can understand the difference very easily with an example, as shown in the picture bellow. Tom goes to school in the morning and then comes back in the afternoon. The total distance he traveled is 10 km, since he walked 5 km going to school, and 5 km getting back. Notice that the direction does not matter for the distance, what matters is how much ground was covered. Distance is therefore a scalar quantity.

But what about the displacement? At the start of the day Tom is at home, and after getting back from school in the afternoon, he is back home again. Since he traveled 5 km to the right (+5 km) and then 5 km to the left (-5 km), in the end the displacement is +5 km — 5 km = 0 km. This means the original and final positions are the same. And because the direction matters for the calculation, displacement is a vector quantity.

# Displacement formula

Definition: The displacement is equal to the distance by Pythagoras’s theorem

∆s = sqrt( (x2 — x1)2 + (y2 — y1) 2 )

Where ∆s is the displacement, x2, y2 the final position and x1, y1 the initial position.

# 1) If the initial and final positions are given

In this case we simply use the formula above.

## Example 1.1: Calculate the displacement of a point that starts at (3,4) and ends at (6,5)

Here we have:

x1 = 3 x2 = 6

y1 = 4 y2 = 5

So, the displacement is:

∆s = sqrt( (x2 — x1)2 + (y2 — y1) 2 ) = sqrt( (6–3)2 + (5–4) 2 ) = 3.162

## Example 1.2: Calculate the displacement of a point that start at (1,5) then passes through (3,7) and ends at (2,4)

Remember than for the displacement we only need the start and end point, which means that the middle point (3,7) is irrelevant. So, we only take the first and the last point:

x1 = 1 x2 = 2

y1 = 5 y2 = 4

So, the displacement is:

∆s = sqrt( (x2 — x1)2 + (y2 — y1) 2 ) = sqrt( (2–1)2 + (4–5) 2 ) = 1.414

# 2) If the velocity and time are given

Sometimes they do not tell us the initial and final point, but instead we get the velocity and time. In this case the formula becomes: ∆s = v*t

## Example 2.1: Calculate the displacement of a car that is traveling at 80 km/h for 45 min.

v = 80 km/h

t = 45 min = 0.75 h

We had to transform the time into hours because the velocity is expressed in hours. The displacement is therefore:

∆s = v*t = 80 km/h * 0.75 h = 60 km

## Example 2.2: Calculate the displacement of a bike that starts at (4,5) and is traveling at 12 km/h for 1 hour. The bike passes through the point (3,6) as well.

Because they do not give us the final point, we cannot calculate the displacement with the square root formula. But we are given the velocity and the time, which means that the initial point and the point that the bike goes through are irrelevant.

v = 12 km/h

t = 1 h

The displacement is therefore:

∆s = v*t = 12 km/h * 1 h = 12 km

# 3) If the velocity, time and also acceleration are given

This is a more general case, where we also have an acceleration. The section before was the special case where the acceleration was 0. However, if there is an acceleration, then the formula becomes: ∆s = v*t + 0.5*a*t2

## Example 3.1: A rocket is traveling at 50 km/h with an acceleration of 1 m/s2. Calculate the displacement of the rocket after 2 hours.

v = 50 km/h

a = 1 m/s2 = 12960 km/h2

t = 2 h

We had to transform the acceleration into km and hours and because the velocity is expressed in those terms. The displacement is therefore:

∆s = v*t + 0.5*a*t2 = 50 km/h * 2 h + 0.5*12960 km/h2*(2 h)2 = 26020 km

## Example 3.2: A spaceship is traveling at 43 km/h with an acceleration of 3 m/s2. Calculate the displacement of the spaceship after 6 hours.

v = 43 km/h

a = 3 m/s2 = 38880 km/h2

t = 6 h

We had to transform the acceleration into hours and km because the velocity is expressed in those terms. The displacement is therefore:

∆s = v*t + 0.5*a*t2 = 43 km/h * 6 h + 0.5*38880 km/h2*(6 h)2 = 700098 km

# Why is displacement important in physics?

If we want to calculate the work done on an object, we have the formula W = F*∆s. We therefore see that only when there is a displacement do we have work. But if the final and initial positions are the same, the displacement is zero, and there is no work. This is one area where displacement is relevant.

Displacement is also useful to know the direction of the final position. If we get a displacement of -2 m, then we know the final position is 2 m to the left of the starting position. We can know this because the displacement is a vector quantity. If we were to compute the distance traveled, we would get a scalar value of 5 m for example, and we would have no way of knowing if the final position is to the right or left from the initial position.