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A particle is moving with $10 \mathrm{~m} / \mathrm{s}$ in a circle of radius 5 m , find out magnitude of average velocity if particle moved by $60^{\circ}$ in 1 sec.
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$5 \mathrm{~m} / \mathrm{s}$
$\begin{aligned} & \text { Given, velocity of particle } v=10 \mathrm{~m} / \mathrm{s} \\ & \text { radius }\left(r^{\prime}\right)=5 \mathrm{~m} \\ & \text { Angular displacement }(\theta)=60^{\circ} \\ & \therefore \quad \theta=60 \times \frac{\pi}{180} \mathrm{rad} \\ & \Rightarrow \quad \theta=\frac{\pi}{3} \mathrm{rad} \\ & \end{aligned}$
When particle moves at an angle of $60^{\circ}$, then according to figure, displacement AB is equal to radius $r$ because it forms an equilateral triangle.
$\therefore$ Average velocity $=\frac{\text { displacement } \mathrm{AB}}{\text { time }}=\frac{\mathrm{r}^{\prime}}{1}=5 \mathrm{~m} / \mathrm{s}$
When particle moves at an angle of $60^{\circ}$, then according to figure, displacement AB is equal to radius $r$ because it forms an equilateral triangle.
$\therefore$ Average velocity $=\frac{\text { displacement } \mathrm{AB}}{\text { time }}=\frac{\mathrm{r}^{\prime}}{1}=5 \mathrm{~m} / \mathrm{s}$
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