91Ó°ÊÓ

The car is traveling at the initial speed$u=30m/s$

The car is traveling on the slope inclined at the angle 10^o.
If the gas runs out, then the car would stop and starts rolling back. Before, it starts rolling the velocity of the car would be $v_f=0m/sec$

Acceleration due to gravity is $g=10m/sec$

The deceleration of the car is given by

localid="1648196205233" $$\begin{gathered} a=g\sin \mathrm{Θ} \\ a=10(\sin 10°) \\ a=1.73m/s^2 \end{gathered}$$

Let the time taken by the car until it starts rolling is$t$.
Write the equation and substitute the known variables
$$\begin{gathered} v=u+at \\ 0=30+(-1.73)t \\ t=\frac{30}{1.73}=17.34sec \end{gathered}$$

Thus, the time taken by a car to stop and roll back is 17.34 sec
Now, let the distance covered by the car is $S$
Using the equation of motion,
localid="1648466936650" $$\begin{gathered} S=ut+\frac{1}{2}a{t}^2 \\ S=30(17.3)+\frac{1}{2}(-1.73){(17.3)}^2 \\ S=519-258.8 \\ S=260.2m \end{gathered}$$

Hence, the distance traveled by car before it rolls back is 260.2 m.