Lesson 60 of 100 | Quantitative Aptitude / Speed, Distance and Time / चाल, दूरी और समय
Trains crossing people, platforms and other trains
Learning outcome
Identify the distance needed for a complete train crossing and combine it with the appropriate relative speed.
Concepts and assumptions
A train’s rear must clear the object; its front reaching it is insufficient. Assume fixed train lengths, straight parallel paths, constant speeds, no stops and sufficient track. Treat people as points.
For a stationary person or pole, time starts when the front reaches the point and ends when the rear passes it. The train moves its length L: time = L ÷ speed.
For a platform of length p, time starts at front entry and ends at rear exit. The front moves p to reach the far end, then another L to bring the rear through. Required distance = L + p.
For a moving person, divide L by relative speed: subtract speeds for same-direction overtaking; add them when approaching in opposite directions.
Two trains completely passing require relative displacement equal to their combined lengths. Opposite-direction timing runs from their fronts meeting until their rears clear; use the speed sum. Same-direction timing runs from the faster front reaching the slower rear until the faster rear passes the slower front; use the speed difference.
These intervals exclude any initial gap. If timing starts earlier, add the specified gap using its stated endpoints. Convert km/h to m/s using × 5/18 before dividing lengths in metres.
Worked examples
Example 1 — A moving person. A 210 m train moves at 63 km/h, passing a person walking in the same direction at 9 km/h. Relative speed = 63 - 9 = 54 km/h = 15 m/s. From the front reaching the person to the rear passing them, time = 210 ÷ 15 = 14 s.
Example 2 — A platform. A 144 m train crosses a 216 m platform at 72 km/h. Speed = 72 × 5/18 = 20 m/s. Required distance = 144 + 216 = 360 m. Complete crossing time = 360 ÷ 20 = 18 s, measured from front entry to rear exit.
Example 3 — Complete overtaking. A 252 m train at 81 km/h overtakes a 168 m train at 54 km/h. Both move in the same direction. Relative speed = 27 km/h = 7.5 m/s. Relative displacement = 252 + 168 = 420 m. From the faster front reaching the slower rear to complete clearance, time = 420 ÷ 7.5 = 56 s.
Common mistakes
A front reaching an endpoint does not mean the whole train has crossed. Do not subtract train lengths during overtaking or add an unstated initial gap. Equal-speed trains cannot complete same-direction overtaking.
Practice questions
- A 198 m train passes a stationary pole completely in 11 s. Find its speed in m/s and km/h.
- A 224 m train crosses a 336 m platform at 63 km/h. Find the time from front entry to rear exit.
- Trains of lengths 175 m and 245 m approach at 54 km/h and 72 km/h. Find the time from their fronts meeting until both completely clear.
- A 156 m train moves at 54 km/h. Its front is 90 m before the near end of a 204 m platform. From this instant, how long until its rear clears the far end?
Worked answers
- The crossing distance is 198 m. Speed = 198 ÷ 11 = 18 m/s = 18 × 18/5 = 64.8 km/h.
- Distance = 224 + 336 = 560 m. Speed = 63 × 5/18 = 17.5 m/s. Time = 560 ÷ 17.5 = 32 s.
- Combined length = 175 + 245 = 420 m. Closing speed = 54 + 72 = 126 km/h = 35 m/s. Time = 420 ÷ 35 = 12 s.
- The front must travel 90 + 204 + 156 = 450 m. Speed = 54 × 5/18 = 15 m/s. Time = 450 ÷ 15 = 30 s.