Answer:
8.4 s
Explanation:
First of all, let's convert the initial angular velocity of the engine from rpm into rad/s:
[tex]\omega_i = 2400 rpm = 2400 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=251.2 rad/s[/tex]
The angular velocity at time t is given by:
[tex]\omega(t)= \omega_0 + \alpha t[/tex]
where
[tex]\alpha=-30 rad/s^2[/tex] is the angular acceleration, which is negative because the engine is slowing down
We want to know how long it takes for the drive to stop turning: this is equivalent of calculating the time t at which the angular velocity becomes zero, [tex]\omega(t)=0[/tex]. Using the equation above, we have:
[tex]0=\omega_i + \alpha t\\t=-\frac{\omega_i}{\alpha}=-\frac{251.2 rad/s}{-30 rad/s^2}=8.4 s[/tex]
