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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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$10 \sqrt{2} m / s$

$10 m / s$

$10 \sqrt{3} m / s$

Zero

answer is B.

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Detailed Solution

$Δ V = \sqrt{V^{2} + V^{2} - 2 V^{2} \cos θ} = 2 V \sin \frac{θ}{2}$

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