#### Answer

(a) The tangential acceleration is $0.966~m/s^2$
(b) The astronaut experiences 14 g's of acceleration.

#### Work Step by Step

(a) We can find the final angular speed as:
$\omega = \frac{2\pi~rad}{1.3~s}$
$\omega = 4.83~rad/s$
We can find the angular acceleration as:
$\alpha = \frac{\omega_f-\omega_0}{30~s}$
$\alpha = \frac{4.83~rad/s-0}{30~s}$
$\alpha = 0.161~rad/s^2$
We can find the tangential acceleration;
$a = \alpha ~r$
$a = (0.161~rad/s^2)(6.0~m)$
$a = 0.966~m/s^2$
The tangential acceleration is $0.966~m/s^2$.
(b) We can find the centripetal acceleration at top speed.
$a_c = \omega^2~r$
$a_c = (4.83~rad/s)^2(6.0~m)$
$a_c = 140~m/s^2$
We can express this acceleration in g's.
$a_c = \frac{140~m/s^2}{9.8~m/s^2} = 14~g's$
The astronaut experiences 14 g's of acceleration.