#### Answer

(a) The angular acceleration is $-101~rad/s^2$
(b) The drill rotates through 50 revolutions as it stops.

#### Work Step by Step

(a) We can convert 2400 rpm to units of rad/s:
$\omega = (2400~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$
$\omega = 80\pi~rad/s$
We can find the angular acceleration.
$\alpha = \frac{\omega-\omega_0}{t}$
$\alpha = \frac{0-80\pi~rad/s}{2.5~s}$
$\alpha = -101~rad/s^2$
The angular acceleration is $-101~rad/s^2$.
(b) $\theta = \omega_0~t+\frac{1}{2}\alpha~t^2$
$\theta = (80\pi~rad/s)(2.5~s)+\frac{1}{2}(-101~rad/s^2)(2.5~s)^2$
$\theta = 313~rad$
We can convert the angle from radians to revolutions.
$\theta = (313~rad)(\frac{1~rev}{2\pi~rad})$
$\theta = 50~rev$
The drill rotates through 50 revolutions as it stops.