A car, moving with a speed of 50kmph, can be stopped by applying brakes after travelling at least a distance of 6 m. If the same car is moving at a speed of 100kmph, the minimum distance covered by the car before being stopped is 

Given that a car moving with

Initial speed $$\begin{gathered} u_1=50kmph=50\times \frac{5}{18}=\frac{125}{9}m/s \\ Brought to rest v_1=0 \\ Let acceleration caused by brakes=-a \\ Dis\tan ce travelled before coming to rest s_1 =6m \\ We know v^2-u=2as \\ 0-{u_1}^2=-2as \\ \therefore {u_1}^2=2a{s}_1\to \left(1\right) \\ It same car with same brakes acceleration=-a \\ Has initial velocity u_2=100kmph=100\times \frac{5}{18}=\frac{250}{9}m/s \\ Brought to rest i.e.v_2=0 \\ Let the dis\tan ce travelled s_2=? \\ We can have \\ {v}^2-u^2=2as \\ 0-{u_2}^2=-2a.s_2 \\ \therefore {u_2}^2=2a{s}_2\to \left(2\right) \\ From \left(1\right) \& \left(2\right) we get \\ \frac{{u_1}^2}{{u_2}^2}=\frac{2a{s}_1}{2a{s}_2} \\ \Rightarrow {\left(\frac{u_1}{u_2}\right)}^2=\frac{s_1}{s_2} \\ {u}_1=\frac{125}{9},u_2=\frac{250}{9},s_1=6m,s_2=? \\ {\left(\frac{{\displaystyle \frac{{\displaystyle \frac{125}{9}}}{250}}}{9}\right)}^2=\frac{6}{s_2} \\ \Rightarrow \frac{1}{4}=\frac{6}{s_2} \\ \therefore s_2=24m \end{gathered}$$