Two trains, each 162-m long, moving in opposite directions, cross each other…

Concept — relative speed. When two bodies move in opposite directions, they approach each other at the sum of their individual speeds; when they move in the same direction, the gap between them changes at the difference of their speeds.

For two moving objects to completely cross each other, the relative displacement must equal the sum of their two lengths. So the time to cross = (sum of the two lengths) ÷ (relative speed).

Unit bridge: 1 m/s = 18/5 km/hr, and 1 km/hr = 5/18 m/s.

Applying this to the given situation:

1. Total length to be cleared = 162 m + 162 m = 324 m.

2. The two trains cross in 12 s, so the relative speed = 324 m ÷ 12 s = 27 m/s.

3. Let the slower train's speed be v. The other train moves twice as fast, so its speed is 2v.

4. The trains move in opposite directions, so the relative speed is the sum of the two speeds: v + 2v = 3v.

5. Therefore 3v = 27 m/s, which gives v = 9 m/s, so the faster train moves at 2v = 18 m/s.

6. Converting to the unit asked for: 18 × 18/5 = 64.8 km/hr, which is the speed of the faster train.

Cross-check, and the slips to avoid:

• At 18 m/s and 9 m/s the gap closes by 27 m every second, so 324 m is cleared in 324 ÷ 27 = 12 s — exactly the crossing time given in the question.

• 18 m/s equals 64.8 km/hr and 9 m/s equals 32.4 km/hr, so the faster train is indeed exactly twice as fast as the slower one, as the question requires.

• Subtracting the two speeds (2v − v = v) is the same-direction rule and does not apply to trains approaching each other; taking only one train's length (162 m in place of 324 m) is the other frequent slip.