Question

A particle moves 3 m north then 4 m east and finally 6 m
south. Calculate the displacement.
Jeth​

Answer 1

Given

A particle moves 3 m north then 4 m east and finally 6 m

TO FIND

displacement of the particle

Explanation

$\sf from\: the\: figure, \\ \sf initial\: point\: of \:the\: particle\: is \:A \\ \sf and\: final \:point\: is \:E$

$\underline{\sf Displacement \: is\: the \:shortest \:distance}$

$\sf here\: AE=S\: is\: the \:displacement$

$DE=CE-CD =6-3=3m\\ AD=BC=4m$

as CD=BA

$\sf now \:from \:\bigtriangleup ADE\\ \sf it \:is\: a\: right \:angled\: triangle$

$AE^2=AD^2+DE^2$

$s ^{2} = {4}^{2} + {3}^{2} \\ = > {s}^{2} = 16 + 9 \\ = > {s}^{2} = 25 \\ = > s = \sqrt{25 } \\ = > s = 5m$

$\huge \boxed{ \sf \: displacement = 5m}$

Answer 2

Solution:

Note: Diagram of this question attach in attachment file.

So,

The particle starts from A and moves northwards to B. It then moves eastwards to C and then southward to D.

Hence, the total distance covered by particle is,

=> d = AB + BC + CD

=> d = 3 + 4 + 6

=> d = 13 m

Now, Firstly what is displacement?

• Displacement is the shortest distance between the initial and final positions.

From the figure it is clearly that displacement is AD.

By using Pythagoras theorem,

=> AD² = AE² + ED²

Here,

=> AE = BC = 4 m

=> EC = AB = 3 m

∴ ED = 6 - EC = 3 m

Now, put the values,

=> AD² = 4² + 3²

=> AD² = 16 + 9

=> AD² = 25

=> AD = 5

Hence, the displacement of particle is 5 m.