Question

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

• A. $0$
• B. $10m/s$
• C. $10\sqrt{3}m/s$
• D. $10\sqrt{2}m/s$

Answer 1

Answer: (B)

$10m/s$

Change in velocity of a particle in circle in given by,

$|\Delta \overline{v}|=\sqrt{v_1^2+v_2^2+2v_1{v}_2\cos (\pi -\theta )}$

Since $[|\overline{v_1}|=|\overline{v_2}|]=v$

$|\Delta \overline{v}|=2v\sin \frac{\theta }{2}$

On putting the values of $vand\theta$
$=(2\times 10)\times \sin \left({30}^o\right)$
$=10m/s$

Hence the change in velocity is $10m/s$