Answer
Time taken $= 8.0$ $seconds$
Work Step by Step
First, to find the angular velocity at the beginning of the $4.0s$ interval, we use the expression:
$\Delta\theta = \omega_{o}t+\frac{1}{2}\alpha t^{2}$
where we have $\Delta\theta=120 rad$, $\alpha=3.0rad/s^{2}$ , $t=4.0s$ (as this is the $4.0s$ interval) and unknown $\omega_{o}$.
So, plugging the values and solving we get:
$120=4\omega_{o}+\frac{1}{2}(3.0)(4^{2})$
$120= 4\omega_{o}+24$
$\omega_{o}=96/4=24$
$\omega_{o}= 24$ $rad/s$
So, we got the angular velocity at the beginning of the $4s$ interval as $24$ $rad/s$.
Now, we need the time taken to reach this interval.
For this we use the expression:
$\omega= \omega_{o}+\alpha t$
Here, we have the unknown time $t$ , $\omega_{o}= 0$ (as it started from rest), $\alpha= 3rad/s^{2}$ , $\omega=24rad/s$ (as the velocity is finally $24rad/s$ that is, at the start of 4s interval).
Plugging the respective values and solving the above expression :
$24=0+3t$
$24/3 = t$
$t= 8.0$ $seconds$
