Question

A particle is moving with constant speed in a circular path. Find the ratio of average velocity from $\theta =0$ to $\theta =\frac{\pi }{2}$ to its instantaneous velocity at $\theta =\frac{\pi }{2}$.

• A. $\frac{\sqrt{2}}{\pi }$
• B. $\frac{\sqrt{2}}{2\pi }$
• C. $\frac{2\sqrt{2}}{\pi }$
• D. $\frac{2\sqrt{2}}{3\pi }$

Answer 1

The correct option is C $\frac{2\sqrt{2}}{\pi }$


We know that the displacement $D=AB=\sqrt{2}r$
Average velocity $v_{\text{avg}}=\frac{D}{t}=\frac{\sqrt{2}r}{t}$
Time taken to describe angle $\theta =\frac{\pi }{2}$,
$t=\frac{\theta }{\omega }=\frac{\theta r}{v}=\frac{\pi r}{2v}$

By putting the value of $t$ in $v_{\text{avg}}$, we get
$v_{\text{avg}}=\frac{2\sqrt{2}v}{\pi }$

Instantaneous velocity $=v$

Thus, the ratio of average velocity to instantaneous velocity $=\frac{2\sqrt{2}}{\pi }$