Question

A particle is displaced 3m towards East , 4m towards North and then 5m vertically upwards. Find the displacement of the particle from the initial position.

Answer 1

Given:

A particle is displaced 3 m towards East

Then, 4 m towards North

Then 5 m vertically upwards

To Find:

Displacement of the particle from the initial position

Diagram:

(In the attachment)

Solution:

• Displacement is the shortest distance between starting and final position
• According to Pythagoras theorem, in a right-angles triangle

$\pink{\boxed{\sf{(Hypotenuse)^{2}=(Perpendicular)^{2}+(Base)^{2}}}}$

$\rule{190}{1}$

In ΔABC

AB= 3 m

BC= 4 m

On applying Pythagoras theorem, we get

$\longrightarrow\rm{(AC)^{2}=(BC)^{2}+(AB)^{2}}$

$\longrightarrow\rm{(AC)^{2}=(4)^{2}+(3)^{2}}$

$\longrightarrow\rm{(AC)^{2}=16+9}$

$\longrightarrow\rm{(AC)^{2}=25}$

$\longrightarrow\rm{AC=\sqrt{25}=5\:m}$

$\rule{190}{1}$

In ΔACD

AC= 5 m

CD= 5 m

On applying Pythagoras theorem, we get

$\longrightarrow\rm{(AD)^{2}=(AC)^{2}+(CD)^{2}}$

$\longrightarrow\rm{(AD)^{2}=(5)^{2}+(5)^{2}}$

$\longrightarrow\rm{(AD)^{2}=25+25}$

$\longrightarrow\rm{(AD)^{2}=50}$

$\longrightarrow\rm{AD=\sqrt{50}=5\sqrt{2}\:m}$

$\rule{190}{1}$

Let d be the displacement of the given particle from the initial position

According to diagram,

AD is the the required displacement of given particle

So,

$\longrightarrow\rm{d=AD}$

$\longrightarrow\rm\green{d=5\sqrt{2}\:m\:towards\:North-East}$

Hence, the displacement of the particle from the initial position is 5√2 m towards North-East.