Question

A train of length $100 m$ travelling at $20 m s^{-1}$ overtakes another of length $200 m$ travelling at $10 m s^{-1}$. The time taken by the first train to pass the second train is

Answer 1

Hint: The length of both the trains are given. The distance travelled by the first train is equal to the sum of the lengths of both the trains. The both trains are travelling in the same direction, the relative velocity is given by the difference of the velocities of both trains. Then, the time is given by the distance divided by the relative velocity.

Complete step-by-step solution:
Given: Velocity of first train $= 20 m s^{-1}$
Velocity of second train $=10 m s^{-1}$
Length of the train = $100 m$.
First train overtakes second train of length $200 m$.
Distance covered by the first train to overtake the second train is equal to the sum of lengths of first and second trains.
Distance $= 100 + 200 = 300 m$
First train overtakes another train, which means both trains are moving in the same direction.
The relative velocity of the first train is given by the difference of velocities of both trains.
Relative velocity of the first train with respect to second train $= 20 – 10 = 10 m s^{-1}$.
Therefore, time needed is the ratio of distance to the relative velocity.
$t = \dfrac{300}{10}$
$\implies t = 30 s$
So, the time taken by the first train to pass the second train is $30 s$.

 Note: If the two bodies are moving in the same direction, then the relative velocity is given by the difference of the velocities of both bodies. If the two bodies are moving in the opposite direction, then the relative velocity is given by the sum of the velocities of both bodies.