Question

Between the two stations, a train uniformly accelerates from rest at first, then moves with constant velocity and finally decelerates uniformly to come to rest. If the ratio of the time taken is $1:8:1$ and the maximum speed attained is $60\text{km/h}$, then what is the average speed over the whole journey?

• A. $48\text{km/h}$
• B. $52\text{km/h}$
• C. $54\text{km/h}$
• D. $56\text{km/h}$

Answer 1

The correct option is C $54\text{km/h}$

Let us divide total journey between two stations into $P$, $Q$ and $R$. Where $P$ is journey when train uniformly accelerates, $Q$ is journey when train moves with constant speed and $R$ is journey when train decelerates uniformly. Also assume the time taken by journey's $P$, $Q$ and $R$ be $t_P$, $t_Q$ and $t_R$ respectively.

As, $t_P:t_Q:t_R=1:8:1$. We can say,
$t_P=t,t_Q=8t$ and $t_R=t$.
As the maximum speed attained is $60km/hr$, we can plot the $v-t$ graph:
Now,
average velocity is given by
$\overline{v}=\frac{\text{total displacement}}{\text{total time}}$

Total displacement is given by area under $v-t$ curve.
$\therefore \text{total displacement}=\frac{1}{2}\times 60\times (8t+10t)\text{total displacement}=30\times 18t$

$\overline{v}=\frac{30\times 18\mathrm{t/}}{10\mathrm{t/}}\Rightarrow \overline{v}=54km/hr$