A particle is moving along a circular path with a constant speed of $10 m s^{- 1}$ . What is the magnitude of the change in velocity of the particle, when it moves through an angle of $60^{0}$ around the centre of the circle?

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$10 \sqrt{2} m / s$

$10 \sqrt{3} m / s$

$10 m / s$

answer is D.

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Detailed Solution

Change in velocity $| Δ \bar{v} | = \sqrt{v_{1}^{2} + v_{2}^{2} + 2 v_{1} v_{2} \cos (π - θ)}$

$= 2 v \sin \frac{θ}{2} (∵ | {\vec{v}}_{1} | = | {\vec{v}}_{2} |) = v = (2 \times 10) \times \sin (30^{0}) = 2 \times 10 \times \frac{1}{2} = 10 m / s$

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