Trains and Platforms

This topic is a specific application of Speed, Distance & Time, where the object in motion is a train, and the focus is on how long it takes to pass a stationary or moving object such as a man, pole, platform, or another train.

Key principle:

Time to cross = Total distance to be covered / Relative speed

2. Key Formulas & Shortcuts

Unit Conversions:

  • 1 km/hr=518 m/s1 \text{ km/hr} = \frac{5}{18} \text{ m/s}
  • 1 m/s=185 km/hr1 \text{ m/s} = \frac{18}{5} \text{ km/hr}

Train Crosses a Pole (or a point object):

  • Time = Length of trainSpeed of train\frac{\text{Length of train}}{\text{Speed of train}}

Train Crosses a Platform (or stationary object with length):

  • Time = Length of train+Length of platformSpeed of train\frac{\text{Length of train} + \text{Length of platform}}{\text{Speed of train}}

Train Crosses a Man Walking (in same or opposite direction):

  • Time = Length of trainRelative speed\frac{\text{Length of train}}{\text{Relative speed}}

    • Relative speed = Subtract if same direction, add if opposite

Two Trains Crossing Each Other:

  • If moving in opposite directions:

    Time=Length of train 1+Length of train 2S1+S2\text{Time} = \frac{\text{Length of train 1} + \text{Length of train 2}}{S_1 + S_2}

  • If moving in same direction:

    Time=Length of train 1+Length of train 2∣S1−S2∣\text{Time} = \frac{\text{Length of train 1} + \text{Length of train 2}}{|S_1 - S_2|}

Train Overtaking or Being Overtaken:

  • Apply relative speed and total distance (combined lengths of the trains or person)

3. Conceptual Tips & Common Mistakes

  • Always convert speeds to the same unit as distance (usually meters per second).

  • Don’t forget to add the platform’s or second train’s length when needed.

  • Be careful with relative speed direction:

    • Same direction → subtract speeds
    • Opposite direction → add speeds

  • For crossing, the entire length of the train must clear the object.

  • Platform length is often hidden in the question—read carefully.

4. Visual Explanation

Diagram 1: Train Crossing a Platform

[========TRAIN========>]  →→→
 ----------------------
|                      |
|      PLATFORM        |
|______________________|

Let train = 180m, platform = 120m, speed = 54 km/hr = 15 m/s
Total length = 300m
Time = 300 / 15 = 20 seconds

Diagram 2: Two Trains Crossing Each Other

When two trains cross head-on, their lengths add, and relative speed increases. Think of them “shrinking the gap” between them faster.

5. Solved Examples

Example 1: Train Passing a Pole

Q: A train 240 m long runs at 60 km/hr. How long does it take to pass a pole?

A:
Speed = 60×10003600=16.67 m/s\frac{60 \times 1000}{3600} = 16.67 \text{ m/s}
Time = 24016.67=14.4\frac{240}{16.67} = 14.4 seconds

Example 2: Train Crossing a Platform

Q: A train 120 m long passes a platform 180 m long in 15 seconds. Find the speed.

A:
Distance = 120 + 180 = 300 m
Speed = 30015=20 m/s=72 km/hr\frac{300}{15} = 20 \text{ m/s} = 72 \text{ km/hr}

Example 3: Two Trains Crossing Opposite Directions

Q: Train A (150m) at 60 km/hr and Train B (100m) at 90 km/hr cross each other. Find time taken.

A:
Relative speed = 60 + 90 = 150 km/hr = 150×10003600=41.67 m/s\frac{150 \times 1000}{3600} = 41.67 \text{ m/s}
Distance = 150 + 100 = 250 m
Time = 25041.67≈6\frac{250}{41.67} \approx 6 seconds

Example 4: Train Passing a Walking Man

Q: A train 120 m long overtakes a man walking at 6 km/hr in 6 seconds. Find the speed of train.

A:
Let speed of train = xx m/s
Speed of man = 6×10003600=1.67 m/s\frac{6 \times 1000}{3600} = 1.67 \text{ m/s}
Relative speed = x−1.67x - 1.67

So,

120x−1.67=6⇒x=21.67 m/s=78 km/hr\frac{120}{x - 1.67} = 6 \Rightarrow x = 21.67 \text{ m/s} = 78 \text{ km/hr}
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