Question

A railway track runs parallel to a road until a turn brings the road to railway crossing. A cyclist rides along the road everyday at a constant speed $20\text{km/hr}$. 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 $10\text{km}$ before the railway crossing. The speed of the train (in $\text{km/hr}$) is $30\text{N}$ ,then $\text{N}$ is

Answer 1

$t_{\text{cycle}}=\frac{10\text{km}}{20\text{km/ h}}=\frac{1}{2}h=30\text{min}$
$\Delta t=5\text{min}=\frac{5}{60}\text{hr}$
Train running as per schedule
So $V_{\text{train}}=\frac{10}{\left(\frac{5}{60}\right)}=\frac{10\times 60}{5}=120{\text{kmh}}^{-1}$