Q. A particle is moving along a circular path with a constant speed of $10 \; ms^{-1}$. What is the magnitude of the change is velocity of the particle, when it moves through an angle of $60^{\circ}$ around the centre of the circle?

Solution:

$\left|\Delta\bar{v}\right| = \sqrt{v_{1}^{2} +v_{2}^{2} +2v_{1}v_{2}\cos\left(\pi-\theta\right)} $
$ = 2v \sin \frac{\theta}{2} \text{since} \left[\left|\bar{v_{1}}\right| = \left|\bar{v_{2}}\right|\right] $
$=\left(2\times10\right)\times\sin\left(30^{\circ}\right) $
$ =10 m/s$

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