Question:
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 \(\mathrm{ms^{-1}}\)) of this person as observed from train B will be close to :
(take the distance between the tracks as negligible)
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 \(\mathrm{ms^{-1}}\)) of this person as observed from train B will be close to :
(take the distance between the tracks as negligible)
(take the distance between the tracks as negligible)
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Convert to m/s (divide km/h by 3.6), take train A direction as positive and subtract velocities.
Updated On: Nov 5, 2026
- \(30.5\ \mathrm{ms^{-1}}\)
- \(29.5\ \mathrm{ms^{-1}}\)
- \(31.5\ \mathrm{ms^{-1}}\)
- \(28.5\ \mathrm{ms^{-1}}\)
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The Correct Option is B
Solution and Explanation
Take the direction of train A as positive.
\(v_A = 36\ \mathrm{km/h} = 10\ \mathrm{m/s}\); train B moves opposite: \(v_B = -72\ \mathrm{km/h} = -20\ \mathrm{m/s}\).
Person walks opposite to train A at \(1.8\ \mathrm{km/h}=0.5\ \mathrm{m/s}\): \(v_P = 10-0.5 = 9.5\ \mathrm{m/s}\).
Velocity of person relative to train B: \(v_P - v_B = 9.5-(-20)=29.5\ \mathrm{m/s}\).
Answer: (2) 29.5 ms-1
\(v_A = 36\ \mathrm{km/h} = 10\ \mathrm{m/s}\); train B moves opposite: \(v_B = -72\ \mathrm{km/h} = -20\ \mathrm{m/s}\).
Person walks opposite to train A at \(1.8\ \mathrm{km/h}=0.5\ \mathrm{m/s}\): \(v_P = 10-0.5 = 9.5\ \mathrm{m/s}\).
Velocity of person relative to train B: \(v_P - v_B = 9.5-(-20)=29.5\ \mathrm{m/s}\).
Answer: (2) 29.5 ms-1
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