#### Answer

(a) $\alpha = 4.0~rad/s^2$
(b) $a_t = 0.12~m/s^2$

#### Work Step by Step

(a) $\omega = (2.0+\frac{1}{2}t^2)~rad/s$
$\alpha = \frac{d\omega}{dt}$
$\alpha = (t)~rad/s^2$
We can find the angular acceleration at t = 4.0 s:
$\alpha = (t)~rad/s^2$
$\alpha = 4.0~rad/s^2$
(b) We can find the tangential acceleration $a_t$ of a tooth on the gear.
$a_t = \alpha ~r$
$a_t = (4.0~rad/s^2)(0.030~m)$
$a_t = 0.12~m/s^2$