Question

A particle moves in a circular path with uniform speed v. When it turns by 90°, find ratio of $\frac{v}{<\overrightarrow{v}>}$?

• A. $\frac{\pi}{\sqrt{2}}$
• B. $\frac{2\pi}{\sqrt{2}}$
• C. $\frac{\pi}{2}$
• D. $2\pi$

Answer 1

Answer: (A)

$\frac{\pi}{\sqrt{2}}$

Answer 2

$|<\overrightarrow{v}>|=\frac{|displacement|}{time}=\frac{\sqrt{2}R}{\frac{\pi R}{v}}=\frac{\sqrt{2}v}{\pi}$
$\frac{v}{<\overrightarrow{v}>}=\frac{v}{(\frac{\sqrt{2}v}{\pi})}=\frac{\pi}{\sqrt{2}}$

So, the correct answer is (A): $\frac{\pi}{\sqrt{2}}$