The driver of a train A travelling at a speed of 60km/h applies brakes and retards the train uniformly. The train stops in 5s. another train B is travelling on the parallel track with a speed of 36 kmph. This driver also applies the brakes and the train retards uniformly. The train B stops in 10 s then distance travelled by each train after the brakes were applied is

$\begin{array}{l}\mathrm{u}=60\mathrm{kmph} \mathrm{u}=36\mathrm{kmph}\\ \mathrm{t}=5s t=10\mathrm{s}\\ \mathrm{v}=0 \mathrm{v}=0\\ \mathrm{v}=\mathrm{u}+ \mathrm{at} \mathrm{v}=\mathrm{u}+ \mathrm{at} \\ 0=\mathrm{u}+ \mathrm{at} 0=\mathrm{u}+\mathrm{at}\\ \mathrm{at} =-\mathrm{u} \mathrm{at} =-\mathrm{u}\\ \mathrm{a}=\frac{-\mathrm{u}}{\mathrm{t}} \mathrm{a}=\frac{-\mathrm{u}}{\mathrm{t}}\end{array}$
$\begin{array}{l}=\frac{60\times 5}{18\times 5} =\frac{-36\times 5}{18\times 10}\\ -\mathrm{a}=\frac{60}{18}\mathrm{m}/{\mathrm{s}}^2 \mathrm{a}=-2\mathrm{m}/{\mathrm{s}}^2\\ {\mathrm{s}}_{\mathrm{A}}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2 {\mathrm{s}}_{\mathrm{B}}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2\\ =\frac{60\times 5}{18}\times 5+\frac{1}{2}\left(\frac{-60}{188}\right)\times 5^2 {\mathrm{s}}_{\mathrm{B}}=\frac{36\times 5}{18}\times 10-\frac{1}{2}\left(1\right)\left(100\right)\\ =\frac{60\times 25}{18}-\frac{30}{18}\times 25 {\mathrm{s}}_{\mathrm{B}}=100-50\\ =\frac{1}{18}(150-750) {\mathrm{s}}_{\mathrm{B}}=50\mathrm{m}\\ {\mathrm{s}}_{\mathrm{A}}=\frac{750}{18}=41.6\mathrm{m}\end{array}$