A car, moving with a speed of $$50 \,km/hr,$$  can be stopped by brakes after at least $$6 \,m.$$  If the same car is moving at a speed of $$100 \,km/hr,$$   the minimum stopping distance is-

Solution :
$$\eqalign{ & {\bf{Case - 1:}} \cr & u = 50 \times \frac{5}{{18}}\,m/s, \cr & v = 0,\,\,\,\,s = 6\,m,\,\,\,\,a = a \cr & {v^2} - {u^2} = 2as \cr & \Rightarrow {0^2} - {\left( {50 \times \frac{5}{{18}}} \right)^2} = 2 \times a \times 6 \cr & \Rightarrow - {\left( {50 \times \frac{5}{{18}}} \right)^2} = 2 \times a \times 6.....(i) \cr & {\bf{Case - 2:}} \cr & u = 100 \times \frac{5}{{18}}\,m/s, \cr & v = 0,\,\,\,\,s = s,\,\,\,\,a = a \cr & \therefore {v^2} - {u^2} = 2as \cr & \Rightarrow {0^2} - {\left( {100 \times \frac{5}{{18}}} \right)^2} = 2 \times a \times s \cr & \Rightarrow - {\left( {100 \times \frac{5}{{18}}} \right)^2} = 2 \times a \times s.....(ii) \cr & {\text{Dividing (i) and (ii) we get}} \cr & \frac{{100 \times 100}}{{50 \times 50}}{\text{ = }}\frac{{2 \times a \times s}}{{2 \times a \times 6}} \cr & \Rightarrow s = 24\,m \cr} $$