If displacement $s=5 \sin (2 t)$, then the velocity at the end of $\frac{\pi}{3} \mathrm{~s}$ is

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Aicharya · Accepted Answer

$s=5 \sin (2 t)$ differentiate w.r.t. ' $t$ ' $ \begin{aligned} & \frac{d s}{d t}=5 \cdot \frac{d}{d t} \sin (2 t) \ & \frac{d s}{d t}=5 \cdot \cos (2 t) \cdot 2=10 \cdot \cos 2 t \end{aligned} $ Velocity when $t=\frac{\pi}{3} \sec =\left.\frac{d s}{d t}ight|_{t=\pi / 3}=10 \cdot \cos 2\left(\frac{\pi}{3}ight)$ $ =10 	imes-\frac{1}{2}=-5 $ Hence, option (4) is correct

If displacement $s=5 \sin (2 t)$, then the velocity at the end of $\frac{\pi}{3} \mathrm{~s}$ is

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