Train A and train B are running on parallel tracks in the opposite directions with speeds of 36 km/hour and 72 km/hour, respectively. A person is walking in train A in the direction opposite to its motion with a speed of 1.8 km/ hour. Speed (in ms –1 ) of this person as observed from train B will be close to :
(take the distance between the tracks as negligible)

Solution

Physics sir · 3 steps

Fix one positive direction, find the person's ground velocity, then subtract train B's velocity from it.

1. Convert to m/s. Multiply each speed by 5/18. Train A: 36 km/h = 10 m s⁻¹. Train B: 72 km/h = 20 m s⁻¹. Person: 1.8 km/h = 0.5 m s⁻¹.
2. Person on ground. Take train A's direction as positive. The person walks backwards inside A, so the velocity relative to the ground is v_P = 10 - 0.5 = 9.5 m s⁻¹. Train B moves the other way, so v_B = -20 m s⁻¹.

   v_(P/B) = v_P - v_B = 9.5 - (-20) = 29.5 m s⁻¹

3. So. An observer in train B sees the person moving at 29.5 m s⁻¹, which is option (b).

The idea to keep. For relative velocity, first find every velocity relative to the ground with signs, then use v_(P/B) = v_P - v_B. Opposite directions add, walking backwards subtracts.

Answer from the NTA final answer key. Solution by PrepWiser.