Question

A particle is moving with a uniform speed $v$ in a circular path of radius $r$ with the centre at $O$. When the particle moves from a point $P$ to $Q$ on the circle such that $? POQ=\theta$, then the magnitude of the change in velocity is

• A. $2v\,sin\left(2\theta\right)$
• B. zero
• C. $2v\,sin\left(\frac{\theta}{2}\right)$
• D. $2v\,cos\left(\frac{\theta}{2}\right)$

Answer 1

Answer: (C)

$2v\,sin\left(\frac{\theta}{2}\right)$

Answer 2

$\therefore$ Change in magnitude of velocity
$|\Delta v | =\sqrt{v^{2}+v^{2}-2 v^{2} \cos \theta}$
$=2 v \sin \frac{\theta}{2}$