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 $54\text{km/hr}$

Step 1, Given data

The ratio of the time taken = 1 : 8: 1

Maximum speed attained = 60 km/h

Step 2, Finding the average speed

Let the train attain maximum velocity in time $t$. Then using 1st eqn of motion,
$v=u+at$

Putting all the values

$60=0+at\Rightarrow at=60$.................(i)

Distance covered till time $t$

$S_1=ut+\frac{1}{2}a{t}^2=0+\frac{1}{2}a{t}^2$

$=\frac{1}{2}at\left(t\right)=\frac{1}{2}\left(60\right)\left(t\right)=30t$..............[From (i)]

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 $=\frac{\text{Total distance}}{\text{Total time}}$

Average speed = $\frac{30t+480t+30t}{10t}$ = $\frac{540t}{10}$ $=54\text{km/hr}$.

Hence the average speed for the whole journey is 54 km/h