A car is moving with speed of 150 km/h and after applying the break it will move 27 m before it stops. If the same car is moving with a speed of one-third the reported speed, then it will stop after travelling ______ m  distance. 

Given, s = 27m, v = 0

$\begin{array}{l}u=150km/h=150X\frac{5}{18}=\frac{125}{3}m/s\\ U\sin g{v}^2-u^2=2as,weget\\ 0-\frac{{125}^2}{9}=2XaX27\Rightarrow a=\frac{-{125}^2}{2X27X9}m/s^2\end{array}$

Now, when car is moving with one – time of its initial speed, we have

$\begin{array}{l}v=0,u=\frac{1}{3}X150km/h\\ u=50X\frac{5}{18}=\frac{125}{9}m/s\\ nowagainu\sin g{v}^2-u^2=2as,weget\\ 0-\frac{{125}^2}{9^2}=2X\left(\frac{-{125}^2}{2X27X9}\right)Xs\\ \Rightarrow s=\frac{{125}^2}{9^2}X\frac{27X9}{{125}^2}\Rightarrow s=3m\end{array}$