Question

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

Answer 1

Change in velocity,
$|\Delta \bar{v} = | \sqrt{v^2_1 + v^2_2 + 2v_1v_2 \,cos(\pi - \theta})$
$ = 2v sin \frac{\theta}{2} \,\,(\because |\vec{v_1}| = |\vec{v_2}|) = v$
$ = ( 2 \times 10)\times sin(30^{\circ}) = 2\times 10 \times \frac{1}{2}$
$ = 10\,m/s$