Question:
A particle is moving along a circular path with a constant speed of \(10\ \mathrm{ms^{-1}}\). What is the magnitude of the change in velocity of the particle, when it moves through an angle of \(60^\circ\) around the centre of the circle?
A particle is moving along a circular path with a constant speed of \(10\ \mathrm{ms^{-1}}\). What is the magnitude of the change in velocity of the particle, when it moves through an angle of \(60^\circ\) around the centre of the circle?
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For rotation of the velocity vector by angle \(\theta\) at constant speed, \(|\Delta\vec{v}| = 2v\sin\frac{\theta}{2}\).
Updated On: Nov 5, 2026
- zero
- \(10\ \mathrm{m/s}\)
- \(10\sqrt{2}\ \mathrm{m/s}\)
- \(10\sqrt{3}\ \mathrm{m/s}\)
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The Correct Option is B
Solution and Explanation
Speed is constant, so the initial and final velocities have equal magnitude \(v = 10\ \mathrm{m/s}\) and differ in direction by \(\theta = 60^\circ\).
\(|\Delta\vec{v}| = 2v\sin\frac{\theta}{2} = 2\times 10\times\sin 30^\circ = 2\times 10\times\frac{1}{2} = 10\ \mathrm{m/s}\).
Answer: (2) \(10\ \mathrm{m/s}\)
\(|\Delta\vec{v}| = 2v\sin\frac{\theta}{2} = 2\times 10\times\sin 30^\circ = 2\times 10\times\frac{1}{2} = 10\ \mathrm{m/s}\).
Answer: (2) \(10\ \mathrm{m/s}\)
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