A car moving with a speed of $$40\,km/h$$   can be stopped after $$2\,m$$  by applying brakes. If the same car is moving with a speed of $$80\,km/h,$$   what is the minimum stopping distance?

Solution :
According to conservation of energy, the kinetic energy of car = work done in stopping the car i.e. $$\frac{1}{2}m{v^2} = Fs$$
where, $$F$$ is the retarding force and $$s$$ is the stopping distance.
For same retarding force,
$$\eqalign{ & s \propto {v^2} \cr & \therefore \frac{{{s_2}}}{{{s_1}}} = {\left( {\frac{{{v_2}}}{{{v_1}}}} \right)^2} = {\left( {\frac{{80}}{{40}}} \right)^2} = 4 \cr & \therefore {s_2} = 4{s_1} = 4 \times 2 \cr & = 8\,m \cr} $$
Alternative
Initial speed of car $$u = 40\;km/h$$
$$ = 40 \times \frac{5}{{18}}\;m/s = \frac{{100}}{9}\;m/s$$
From 3rd equation of motion,
$$\eqalign{ & {v^2} = {u^2} - 2as \cr & \Rightarrow 0 = {\left( {\frac{{100}}{9}} \right)^2} - 2 \times a \times 2 \cr & 4a = \frac{{100 \times 100}}{{81}} \cr & \Rightarrow a = \frac{{2500}}{{81}}\;m/{s^2} \cr} $$
Final speed of car $$ = 80\;km/h$$
$$ = 80 \times \frac{5}{{18}} = \frac{{200}}{9}\;m/s$$
Suppose car stops for a distance $${s'}.$$ Then
$$\eqalign{ & {v^2} = {u^2} - 2as' \cr & 0 = {\left( {\frac{{200}}{9}} \right)^2} - 2 \times \frac{{2500}}{{81}}s' \cr & \Rightarrow s' = \frac{{200 \times 200 \times 81}}{{9 \times 9 \times 2 \times 2500}} = 8\;m \cr} $$