Question

A railway track runs parallel to a road until a turn brings the road to railway crossing. A cyclist rides along the road every day at a constant speed \(20~km/hr\). He normally meets a train that travels in same direction at the crossing. One day he was late by \(25\) minutes and met the train \(10~km\) before the railway crossing. Find the speed of the train.

Answer 1

Draw a labelled diagram according to given condition.
When cyclist late by \(25\) minutes:

Find the speed of train.
According to given condition,
If on day the man is \(25~min\) late he will reach \(B\) at time \(t+25\).
Time taken by the cyclist to cover \(10~km\)
\(t_\text{cycle}=\dfrac{10~km}{20~km/h}=\dfrac{1}{2}~h=30\)min

Hence, the time of the cyclist at position \(C\) will be
\(t+25-30=(t-5)~min\)

The train is running right time it will reach at time \(t\). Hence, time taken by train to reach from \(C\) to \(B\) is \(5~min\).
It means the train will take \(5~min\) to cover \(10~km\).
\(\Delta t=5~min=\dfrac{5}{60}~hr\)
Train running as per schedule.

So,

\(v_{train}=\dfrac{10}{\dfrac{5}{60}}=\dfrac{10\times60}{5}=120~km/h\)