Question

A particle starting from an initial point O covers the diastance of 10 kilometres towards north and 20 km towards east and then it comes back to its starting point O. Find the displacement of the particle​

Answer 1

Topic :- Motion in straight line

$\maltese\: \underline{\underline{\textsf{ \textbf{AnsWer:}}}}\:\maltese$

A particle starting from initial point "O" and covers a distance of 10 km towards north and again the particle moves towards east and covers a distance of 20 km and again it comes back to its initial point "O".

Let,

• OA = 10 km
• AB = 20 km
• Displacement (OB) = ?

By Pythagoras theorem we can find the Displacement of the particle:

$\longrightarrow\:\:\sf (Hypotenuse)^2=(Perpendicular)^2 + (Base)^2$

$\longrightarrow\:\:\sf (OB)^2 = (OA)^2 + (AB)^2$

$\longrightarrow\:\:\sf (OB)^2 = (10)^2 + (20)^2$

$\longrightarrow\:\:\sf (OB)^2 =100 + 400$

$\longrightarrow\:\:\sf (OB)^2 =500$

$\longrightarrow\:\:\sf OB = \sqrt{500}$

$\longrightarrow\:\:\sf OB = \sqrt{5 \times 100}$

$\longrightarrow\:\:\sf OB = \sqrt{5 \times 10 \times 10}$

$\longrightarrow\:\:\sf OB = 10\sqrt{5 }$

$\longrightarrow\:\:\sf OB = 10 \times 2.23$

$\longrightarrow\:\: \underline{ \boxed{\sf OB = 22.3 \: km}}$

Hence,the displacement (OB) of the particle is 22.3 km.

Answer 2

answer = 22.3 km

( explanation is in the attachment)

hope it helps you ☺☺