# Displacement Formula

Displacement is calculated as the shortest distance between starting and final point which prefers straight-line path over curved paths.

Suppose a body is moving in two different directions x and y then Resultant Displacement will be

$S\,&space;=\,&space;\sqrt{x^{2}+y^{2}}$

It gives the shortcut paths for the given original paths.

$S\,&space;=\,&space;vt$

$S\,&space;=\,&space;\frac{1}{2}\,&space;\left&space;(&space;u&space;+v\right)t$

$S\,&space;=\,&space;ut\,&space;+\,&space;\frac{1}{2}at^{^{2}}$

Here,

u = Initial velocity

v = final velocity

a = acceleration

t = time taken.

## Solved Examples

Problem 1: The path distance from the garden to a school is 5Â m west and then 4Â m south. A builder wants to build a short distance path for it. Find the displacement length of the shortest path.

Solution:

Given: Distance to the west x = 5Â m

Distance to the south y = 4Â m.

Displacement is given by

$S\,&space;=\,&space;\sqrt{x^{2}+y^{2}}$

$s=\sqrt{5^{2}+4^{2}}$

s = 6.403Â m.

The builder can build a path for displacement length of 6.7 m.

Question 2:A girl walks from the corridor to the gate she moves 3Â m to the north opposite to her house then takes a left turn and walks for 5Â m, then she takes right turn and moves for 6Â m and reaches the gate. What is the displacement, magnitude, and distance covered by her?

Solution:

Total distance travelled d = 3Â m + 5Â m + 6Â m = 14Â m.

A magnitude of the displacement can be obtained by visualizing the walking. The actual path from A to B as 3Â m then from B to D as 5Â m and finally from D to E as 6Â m.

So, the magnitude of the resultant displacement is

$$\begin{array}{l}\left | S \right |=\sqrt{AC^2+CE^2}\end{array}$$

From figure AC = AB + BC = 3Â m + 6Â m = 9Â m

BD = CE = 5Â m

|S| =âˆš92+52 = 10.29 m.

The direction of Resultant displacement is South East.