# 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 garden to a school is 5m west and then 4m south. A builder wants to build a short distance path for it. Find

Solution:

Given: Distance to the west x = 5m

Distance to the south y = 4m.

Displacement is given by S = x2+y2−−−−−−√x2+y2

= 52+42−−−−−−√52+42

= 6.70 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 3m to the north opposite from her house then takes a left turn and walks for 5m. then she takes right turn and moves for 6m and reaches the gate. What is the displacement, magnitude, and distance covered by her?

Solution:

Total distance traveled d = 3m + 5m + 6m = 14m.

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

SSo, the magnitude of resultant displacement is |S| = AC2+CE2−−−−−−−−−−√AC2+CE2

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

BD = CE = 5m

|S| = 92+52−−−−−−√92+52 = 10.29 m.

The direction of Resultant displacement is South East.