Question

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

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

Answer 1

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

Let the train attain maximum velocity in time $t$. The using 1st eqn of motion,
$v=u+at=0+at=60$
Distance covered till time $t$
$S_1=ut+\frac{1}{2}a{t}^2=0+\frac{1}{2}a{t}^2$
$=\frac{1}{2}at\left(t\right)=\frac{1}{2}\left(60\right)\left(t\right)=30t$

If the time taken to accelerate is $t$, time taken in travelling with uniform velocity is $8t$.
Distance covered $S_2=60\times 8t=480t$

Since same time $t$ is taken in decelerating from max. velocity to rest, by symmetry, we can say
$S_3=S_1=30t$

Average speed $=\frac{\text{Total distance}}{\text{Total time}}$
$=\frac{S_1+S_2+S_3}{t+8t+t}=\frac{540t}{10t}$
$=54\text{km/hr}$