Answer
(a) The magnitude of deceleration is $58~m/s^2$ which is 5.9$g$.
(b) t = 12 ms
Work Step by Step
(a) We can calculate the rate of deceleration for the hip.
$a = \frac{v^2-v_0^2}{2x} = \frac{(1.3~m/s)^2-(2.0~m/s)^2}{(2)(0.020~m)} = -58~m/s^2$
The magnitude of deceleration is $58~m/s^2$
$a = \frac{58~m/s^2}{9.80~m/s^2} = 5.9 g$
(b) $t = \frac{v-v_0}{a} = \frac{1.3~m/s-2.0~m/s}{-58~m/s^2} = 0.012~s = 12~ms$
