Question

A train starts from rest and for the first kilometer moves with constant acceleration. For the next 3 kilometers it has constant velocity and for another two kilometers it moves with constant retardation to come to rest after a total time of 10 minutes. The maximum velocity of the train is

• A. 18 km / h
• B. 36 km / h
• C. 54 km / h
• D. 39 km / h

Answer 1

Answer: (C)

Let $v$ be the maximum velocity acquired by train while accelerating for time $t_{1}$, covering distance $1 \,km$. Therefore,
$1=\left(\frac{0+v}{2}\right) t_{1} \text { or } 2=v t_{1}\,\,\,\,\,\,\dots(i)$
If $t_{2}$ is the time for uniform motion, then $3=v t_{2}\,\,\,\,\,\dots(ii)$
If $t_{3}$ is the time for retarded motion, then
$2=\left(\frac{v+0}{2}\right) \times t_{3}$ or $4=v t_{3}\,\,\,\,\,\,\dots(iii)$
Adding (i), (ii) and (iii), we get
$ 2+3+4=v\left(t_{1}+t_{2}+t_{3}\right)$
$=v \times \frac{10}{60}$ or $v=54 \,km / h $