Answer
the distance travelled by the car before it starts to roll back $ =264.7m$
Work Step by Step
Intial velocity of the car $ v_i = 30m/s$
Final velocity of the car $ v_f =0$
acceleration of the car $a = -gsin(10) = -1.7m/s^2$
let $s$ be the distance travelled by the car before it starts to roll back. Thus we have
$ 2as =v^2 -u^2$
$ 2* (-1.7)* s = 0 -30^2$
$ s = \frac{900}{3.4}$
$s = 264.7 m$
