Time, speed, distance — relative speed, average speed, train problems

medium CBSE 3 min read

Question

A train 150 m long passes a platform 250 m long in 20 seconds. Find the speed of the train. If another train 200 m long is coming from the opposite direction at 72 km/h, how long will they take to cross each other?


Solution — Step by Step

When a train passes a platform, it must cover its own length plus the platform length.

Total distance =150+250=400= 150 + 250 = 400 m

Speed =40020=20= \frac{400}{20} = 20 m/s =20×185=72 km/h= 20 \times \frac{18}{5} = \mathbf{72 \text{ km/h}}

When two trains move in opposite directions, their relative speed = sum of speeds.

Second train speed = 72 km/h = 20 m/s

Relative speed =20+20=40= 20 + 20 = 40 m/s

Total distance to cover = sum of both train lengths =150+200=350= 150 + 200 = 350 m

Time =35040=8.75 seconds= \frac{350}{40} = \mathbf{8.75 \text{ seconds}}


Why This Works

graph TD
    A["TSD Problem Type"] --> B["Train passing a pole/person"]
    A --> C["Train passing a platform/bridge"]
    A --> D["Two trains crossing each other"]
    A --> E["Average speed problem"]
    B --> F["Distance = train length"]
    C --> G["Distance = train length + platform length"]
    D --> H["Distance = sum of both lengths"]
    D --> I["Same direction: relative speed = difference"]
    D --> J["Opposite direction: relative speed = sum"]
    E --> K["Avg speed = total distance / total time"]

The fundamental formula is Distance=Speed×Time\text{Distance} = \text{Speed} \times \text{Time}. For train problems, the tricky part is always identifying the correct distance. A train is not a point — it has length. So “passing” something means the entire train has to go past it, from engine to last coach.

For relative speed: imagine you are sitting on one train. A train coming toward you appears faster (speeds add). A train going the same way appears slower (speeds subtract).


Alternative Method

Unit conversion shortcut: To convert km/h to m/s, multiply by 5/185/18. To convert m/s to km/h, multiply by 18/518/5. This comes from: 1 km/h=10003600=5181 \text{ km/h} = \frac{1000}{3600} = \frac{5}{18} m/s.

For average speed when distances are equal: if you travel distance dd at speed v1v_1 and return at speed v2v_2, the average speed is NOT (v1+v2)/2(v_1 + v_2)/2. It is the harmonic mean: 2v1v2v1+v2\frac{2v_1 v_2}{v_1 + v_2}. This is a very common trap in board exams.


Common Mistake

Using distance = train length when the train crosses a bridge/platform. The train must cover its own length PLUS the bridge length. Students often write only the bridge length as the distance, forgetting that the last coach also needs to clear the bridge. Similarly, when a train passes a person (or pole), the distance is the train’s length (not zero), because the entire train needs to pass that point.

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