A car moving at a speed v is stopped by a retarding force F in a distance s. If the speed of the car was 3v, the force needed to stop it within the same distance s will be

Let m be the mass of the car and a be the deceleration produced by force F, then F= ma, where a is given by $2\mathrm{as}={\mathrm{v}}^2\text{ or }\mathrm{a}=\frac{{\mathrm{v}}^2}{2\mathrm{s}}$.

Therefore, $\mathrm{F}=\frac{{\mathrm{mv}}^2}{2\mathrm{s}}.$

$\text{Thus }\mathrm{F}\propto {\mathrm{v}}^2$. 

If v is increased by 3 times, F will increase by 9 times.