Question

A man moves 9m North and 16m East then find the displacement of the man from start

Answer 1

Given:

A man moves 9 m North

Then he moves 16 m East

To Find:

Displacement of the man from start

Diagram:

(In the attachment)

Solution:

We know that,

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

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

$\rule{190}{1}$

In ΔABC

AB= 9 m

BC= 16 m

On applying Pythagoras theorem, we get

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

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

$\longrightarrow\rm{(AC)^{2}=256+81}$

$\longrightarrow\rm{(AC)^{2}=337}$

$\longrightarrow\rm{AC=\sqrt{337}}$

$\longrightarrow\rm{AC=18.36\:(approx)}$

$\rule{190}{1}$

Let d be the displacement of the man from the start

According to diagram,

AC is the the required displacement of man

So,

$\longrightarrow\rm{d=AC}$

$\longrightarrow\rm\green{d=18.36\:m\:towards\:North-East}$

Hence, the displacement of man from the start is 18.36 m towards North-East.