Train A and train B are running on parallel tracks in the opposite directions with speed of $36\ \text{km hour}^{-1}$ and $72\ \text{km hour}^{-1}$, respectively. A person is walking in train A in the direction opposite to its motion with a speed of $1.8\ \text{km hour}^{-1}$. Speed (in $\text{m s}^{-1}$) of this person as observed from train B will be close to: (take the distance between the tracks as negligible)
Answer: (A) $29.5\ \text{m s}^{-1}$
Take the direction of train A as positive. In m/s: $v_A = 10$, $v_B = -20$ and the person's speed relative to A is $0.5$ (backwards).
Velocity of the person relative to the ground: $10 - 0.5 = 9.5\ \text{m s}^{-1}$.
Relative to train B: $9.5 - (-20) = 29.5\ \text{m s}^{-1}$.
Solution by Sreeraj P, M.Sc Physics