Question

a train moving at 108km/hr stops in 2 minutes.calculate the retardation and distance covered before stopping.

Answer 1

$$\begin{gathered} Initialvelocityu=108km/h=30m/s \\ timetakent=2min=120s \\ finalvelocityv=0 \\ accelerationis \\ a=\frac{v-a}{t} \\ a=\frac{0-30}{120}=-0.25m/s^2 \\ Dis\tan cetravelled \\ s=ut+\frac{1}{2}a{t}^2 \\ s=30\times 120-\frac{1}{2}\times 0.25\times 120\times 120 \\ s=3600-1800=1800m=1.8km \end{gathered}$$