Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) The tangential acceleration is $0.966~m/s^2$ (b) The astronaut experiences 14 g's of acceleration.
(a) We can find the final angular speed. $\omega = \frac{2\pi~rad}{1.3~s}$ $\omega = 4.83~rad/s$ We can find the angular acceleration. $\alpha = \frac{\omega_f-\omega_0}{30~s}$ $\alpha = \frac{4.83~rad/s-0}{30~s}$ $\alpha = 0.161~rad/s^2$ 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.