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(C) The angular displacement is given by $	heta = \frac{t^{2}}{20} + \frac{t}{5}$.Angular velocity $\omega$ is the rate of change of angular displace

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$3$

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(C) The angular displacement is given by $	heta = \frac{t^{2}}{20} + \frac{t}{5}$.Angular velocity $\omega$ is the rate of change of angular displacement with respect to time,given by $\omega = \frac{d	heta}{dt}$.Differentiating $	heta$ with respect to $t$:$\omega = \frac{d}{dt} \left( \frac{t^{2}}{20} + \frac{t}{5} ight) = \frac{2t}{20} + \frac{1}{5} = \frac{t}{10} + \frac{1}{5}$.At $t = 4 \ s$,the angular velocity $k$ is:$k = \left( \frac{4}{10} + \frac{1}{5} ight) = \frac{4}{10} + \frac{2}{10} = \frac{6}{10} = 0.6 \ rad/s$.We need to find the value of $5k$:$5k = 5 	imes 0.6 = 3$.

$A$ particle starts from rest and its angular displacement (in radians) is given by $\theta = \frac{t^{2}}{20} + \frac{t}{5}$. If the angular velocity at the end of $t = 4 \ s$ is $k$,then the value of $5k$ is

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