Question 308 of 308

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$10 \sqrt{3} \mathrm{~m} / \mathrm{s}$BzeroC$10 \sqrt{2} \mathrm{~m} / \mathrm{s}$D$10 \mathrm{~m} / \mathrm{s}$

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Answer: (D) $10 \mathrm{~m} / \mathrm{s}$

Change in velocity,
$$\begin{array}{l} |\Delta \overline{\mathrm{v}}|=\sqrt{\mathrm{v}_{1}^{2}+\mathrm{v}_{2}^{2}+2 \mathrm{v}_{1} \mathrm{v}_{2} \cos (\pi-\theta)} \\ =2 \mathrm{v} \sin \frac{\theta}{2} \quad\left(\because\left|\overrightarrow{\mathrm{v}}_{1}\right|=\left|\overrightarrow{\mathrm{v}}_{2}\right|\right)=\mathrm{v} \\ =(2 \times 10) \times \sin \left(30^{\circ}\right)=2 \times 10 \times \frac{1}{2} \\ =10 \mathrm{~m} / \mathrm{s} \end{array}$$