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# Displacement Formula

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Last updated date: 08th Sep 2024
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## An Introduction to Displacement

Displacement is the change in the position of any point or object. It has a direction and magnitude, so it is a vector quantity. It is shown with an arrow that points from the starting position to the final position. For example, an object moved from position A to position B, then the position of the object is changed. The shortest distance from position A and position B is called displacement. In this article, we will learn about the formula for displacement.

### Displacement Formulas – How to Calculate Displacement

We find the displacement by determining the object’s initial and final position. The displacement formula is given by:

Δx  = xf − xi….. (1)

where,

xf is the final position,

xi is the initial position, and

Δx is the change in the position of the object.

Let’s say, I kept a golf ball on the origin or the 0 points, as shown in diagram A:

Now, I move the ball by 3 points to the right, as shown in diagram B.

Since the ball has moved, which means it has made a displacement, as we can see its initial position was 0 at A and the final is 3 metres at point B, so its displacement can be calculated by using equation 1, which is:

Δx  = xf − xi = 3 − 0 = + 3 m, i.e., to the right.

So, the ball made a displacement of 3 meters from A to B.

Now, the ball moves by − 4 points or − 4 m to point C.

So, the final position now becomes an initial one and final, xi = − 4  m. So, we can find the displacement AC by using eq (1) as:

Δx  = xf − xi = − 4 − (+ 3) = − 7 m, i.e., to the left.

However, we must remember that displacement doesn’t depend on the position of the origin, as we saw in the diagram, it only depends on the difference between the two positions.

Here,

x1  = x-coordinate of point A, i.e., the initial position of an object along the x-axis

x2 = x-coordinate of point B, i.e., the final position of an object along the x-axis

The horizontal distance, i.e., along the x-axis = (x2 − x1)

y1 = y-coordinate of point A, i.e., the initial position of an object along the y-axis

y2 = y-coordinate of point B, i.e., the final position of an object along the y-axis

The vertical distance, i.e., along the y-axis = (y2 − y1)

We know that displacement is the shortest distance between two points; in a two-dimensional system, a distance formula can be used to find the shortest distance between two points, i.e., P (x1, y1) and Q (x2, y2), which is given as:

$S = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

If an object is moving with constant velocity, then

Displacement = velocity x time

If an object is moving with constant acceleration, then the equation of third law of motion is used to find displacement:

S = ut + ½ at2

$S = \frac{v^{2} - u^{2}}{2a}$

where v = final velocity, u = Initial velocity, s = displacement, and a = acceleration.

### Solved Examples on Displacement Formula

Question 1: Find the displacement covered by an object which accelerates from rest to 60 m/s in 3s.

Solution: Initial velocity = 0 and final velocity = 60 m/s.

Time taken = 3s.

Therefore, acceleration = 60/3 = 20m/s2.

Displacement (S) = ut + ½ at2

= 90 m

Question 2: Find the displacement of an object if it moves with an acceleration of 4 m/s2 for 5s, where the initial velocity was 2 m/s.

Solution: Given initial velocity = 2 m/s, acceleration a  = 4 m/s2 , and t =  5s, so the displacement will be:

Displacement =$\frac{v^{2} - u^{2}}{2a}$

Also, we know that

$a = \frac{v - u}{t}$ => 4 = $\frac{v - 2}{5}$ =>   v = 22

$S = \frac{(22)^{2} - (2)^{2}}{2 \times 4} = \frac{484 - 4}{8} = 60 m$

Question 3:  A Person’s office is 5 m to the west and 4m south of his house.  He wants to take the shortest distance. Determine the shortest distance he covered to reach his office early.

Solution: The shortest distance can be determined by using the pythagoras formula:

$\sqrt{(5)^{2} + (4)^{2}} = \sqrt{25 + 16} = \sqrt{41}$ ~ 6.44 m

Question 4: Mithila throws the cricket ball 30 feet east for her pet. Her pet catches the ball and takes it past you, and you are standing 38 feet to the west of where Mithila is. What is the displacement of the cricket ball?

Solution: Initial position = 0 km and the final position is calculated by 38 − 30 = − 8 km

Now, using the formula, Δx  = xf − xi = − 8 − 0 = − 8 m

D = 8 m of Mithila’s initial position.

### Conclusion

The article is very useful for the students to understand the basic concept of displacement. The solved examples will help students to learn how to apply these formulas in numerical calculations.

## FAQs on Displacement Formula

1. What is the displacement formula?

The displacement between two positions of an object is calculated by measuring the shortest distance between them. The value of displacement can either be zero, negative, or positive. In Physics, we can calculate displacement by calculating the distance between the initial position and the final position of the object. In Physics, often displacement is referred to as the variable s. The displacement formula is as follows: s = sf – sf. Where notations sf is used for the final position and si is used for the initial position.

2. What are the differences between distance and displacement?

The distance is defined as the total path between two points. Distance is a positive quantity whose value can not be negative. Distance is a scalar quantity.

It does not decrease with time. It is never less than the value of the displacement.

The displacement is the shortest distance between the initial or final position or the shortest distance between any two points. Displacement can be zero, negative, or positive depending on the context. Displacement is a vector quantity. It can be decreased or increased with time.