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(D) For the first $5 \ s$: $\omega = \omega_0 + \alpha t = 0 + (0.4)(5) = 2 \ rad \ s^{-1}$.The angular displacement $	heta_1$ is given by $\omega^2

Vedclass · Accepted Answer

$210$

Vedclass · Answer

(D) For the first $5 \ s$: $\omega = \omega_0 + \alpha t = 0 + (0.4)(5) = 2 \ rad \ s^{-1}$.The angular displacement $	heta_1$ is given by $\omega^2 - \omega_0^2 = 2\alpha	heta_1$.$	heta_1 = \frac{2^2 - 0^2}{2(0.4)} = \frac{4}{0.8} = 5 \ rad$.For the next $30 \ s$,the angular velocity is constant at $2 \ rad \ s^{-1}$.$	heta_2 = \omega t = 2 	imes 30 = 60 \ rad$.For the final deceleration phase: $\omega_0 = 2 \ rad \ s^{-1}$,$\omega = 0$,$\alpha = -0.4 \ rad \ s^{-2}$.$	heta_3 = \frac{\omega^2 - \omega_0^2}{2\alpha} = \frac{0^2 - 2^2}{2(-0.4)} = \frac{-4}{-0.8} = 5 \ rad$.Total angular displacement $	heta = 	heta_1 + 	heta_2 + 	heta_3 = 5 + 60 + 5 = 70 \ rad$.The linear distance $l = r	heta = 3 	imes 70 = 210 \ m$.

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