DEFINITION
One-Line Definition
Problems on Trains calculate the time taken by a moving train to cross stationary objects (poles, platforms, bridges) or moving objects (other trains, people) using distance and relative speed.
Key Rule: Time = Total Distance / Effective Speed
CORE INTUITION 🚂
Train Length Mental Model
A train is not a point object; it has length!
- • Crossing a Pole/Person → Distance = Train Length
- • Crossing a Platform/Bridge → Distance = Train Length + Platform Length
- • Crossing Another Train → Distance = Length A + Length B
The entire length of the train must clear the object for the crossing to be complete!
💡 WHY THIS CONCEPT MATTERS & REAL-LIFE APPLICATIONS
Train problems are among the most frequently tested applications of Time, Speed & Distance. Click on any connected topic to jump directly to its Vault page:
Where Are Train Concepts Used?
🚉 Railway Scheduling
🚚 Transportation Logistics
🚥 Traffic Signal Timing
📦 Cargo Tracking
📐 GOLDEN FORMULAS & CONVERSIONS
GOLDEN TRAIN CROSSING FORMULA
Time = Total Distance / Relative Speed
Crossing Time (seconds) = (Length of Train + Length of Object) / Relative Speed (m/s)
Speed Unit Conversions
km/h to m/s
1 km/h = 5 / 18 m/s
Multiply speed by 5/18
m/s to km/h
1 m/s = 18 / 5 km/h
Multiply speed by 18/5
Distance & Relative Speed Matrix
| Scenario | Distance Covered | Effective Speed |
|---|---|---|
| Train crossing a Pole / Tree / Stationary Person | L (Length of Train) | S (Speed of Train) |
| Train crossing a Platform / Bridge / Tunnel | L1 + L2 (Train + Platform Length) | S (Speed of Train) |
| Two Trains in OPPOSITE directions | L1 + L2 (Length A + Length B) | S1 + S2 (Add Speeds) |
| Two Trains in SAME direction | L1 + L2 (Length A + Length B) | |S1 - S2| (Subtract Speeds) |
📝 SOLVED EXAMPLES (LEVEL 0 TO ADVANCED)
EASY • EXAMPLE 1
A train 180 meters long crosses a pole in 9 seconds. Find its speed in km/h.
Solution:
Distance = 180 meters (Length of Train) | Time = 9 seconds
Speed in m/s = 180 / 9 = 20 m/s
Convert to km/h = 20 × (18 / 5) = 4 × 18 = 72 km/h
Answer: 72 km/h
MEDIUM • EXAMPLE 2
A train 150 meters long crosses a 250-meter platform in 20 seconds. Find its speed in km/h.
Solution:
Total Distance = Train Length + Platform Length = 150 + 250 = 400 meters
Time = 20 seconds
Speed in m/s = 400 / 20 = 20 m/s
Convert to km/h = 20 × (18 / 5) = 72 km/h
Answer: 72 km/h
HARD • EXAMPLE 3
Train A is 120m long and travels at 54 km/h. Train B is 180m long and travels at 36 km/h in the opposite direction. Find the time taken to cross each other.
Solution:
Convert Speeds to m/s: S1 = 54 × (5/18) = 15 m/s | S2 = 36 × (5/18) = 10 m/s
Opposite Directions => Relative Speed = S1 + S2 = 15 + 10 = 25 m/s
Total Distance = Length A + Length B = 120 + 180 = 300 meters
Time = Total Distance / Relative Speed = 300 / 25 = 12 seconds
Answer: 12 seconds
⚠️ COMMON MISTAKES TO AVOID
❌ Mistake 1: Ignoring Platform / Bridge Length
When crossing a platform, bridge, or tunnel, the distance is NOT just the train length—it is Train Length + Platform Length!
❌ Mistake 2: Mixing Units (km/h vs m/s)
Never divide distance in meters by speed in km/h! Always convert km/h to m/s by multiplying by 5/18 before calculating time.
❌ Mistake 3: Subtracting Speeds in Opposite Directions
Opposite directions mean trains approach faster → ADD speeds! Same direction means SUBTRACT speeds.
🚀 CAT & MBA CET SHORTCUTS
⚡ Shortcut 1: Quick 5/18 Speed Multiples
Memorize key speed equivalences:
• 18 km/h = 5 m/s | 36 km/h = 10 m/s | 54 km/h = 15 m/s
• 72 km/h = 20 m/s | 90 km/h = 25 m/s | 108 km/h = 30 m/s
• 18 km/h = 5 m/s | 36 km/h = 10 m/s | 54 km/h = 15 m/s
• 72 km/h = 20 m/s | 90 km/h = 25 m/s | 108 km/h = 30 m/s
⚡ Shortcut 2: Pole vs Platform Difference Rule
If a train crosses a pole in T1 seconds and a platform of length P in T2 seconds:
• Speed of Train = P / (T2 - T1) m/s
• Length of Train = Speed × T1 = [P / (T2 - T1)] × T1
• Speed of Train = P / (T2 - T1) m/s
• Length of Train = Speed × T1 = [P / (T2 - T1)] × T1
🎯 PRACTICE QUESTIONS
QUESTION 1 • BASIC
A train 240 meters long crosses a pole in 12 seconds. Find its speed in m/s and km/h.
Speed = 240 / 12 = 20 m/s.
In km/h = 20 × (18 / 5) = 72 km/h.
In km/h = 20 × (18 / 5) = 72 km/h.
QUESTION 2 • BASIC
A train 180 meters long crosses a 120-meter platform in 15 seconds. Find its speed in km/h.
Total Distance = 180 + 120 = 300 meters.
Speed in m/s = 300 / 15 = 20 m/s.
In km/h = 20 × (18 / 5) = 72 km/h.
Speed in m/s = 300 / 15 = 20 m/s.
In km/h = 20 × (18 / 5) = 72 km/h.
QUESTION 3 • MODERATE
Two trains 150 meters and 250 meters long travel in opposite directions at 45 km/h and 63 km/h. Find the time taken to cross each other.
Relative Speed = 45 + 63 = 108 km/h.
Convert to m/s = 108 × (5 / 18) = 30 m/s.
Total Distance = 150 + 250 = 400 meters.
Time = 400 / 30 = 13.33 seconds (40/3 s).
Convert to m/s = 108 × (5 / 18) = 30 m/s.
Total Distance = 150 + 250 = 400 meters.
Time = 400 / 30 = 13.33 seconds (40/3 s).
QUESTION 4 • ADVANCED
A train crosses a pole in 10 seconds and a 300-meter platform in 25 seconds. Find the length and speed of the train.
Use Shortcut: Speed = Platform Length / (T2 - T1) = 300 / (25 - 10) = 300 / 15 = 20 m/s (72 km/h).
Length of Train = Speed × T1 = 20 × 10 = 200 meters.
Length of Train = Speed × T1 = 20 × 10 = 200 meters.
❓ FREQUENTLY ASKED QUESTIONS
Q: Why do we add the platform length when a train crosses a platform?
Because a train is an extended object with length. For the train to completely cross a platform, the front of the train must cover the platform length and the tail of the train must clear the platform, making total distance = Train Length + Platform Length.
Q: When do we use relative speed in train problems?
Whenever two objects (two trains, or a train and a person/car) are moving simultaneously. Use sum of speeds (S1 + S2) when moving in opposite directions, and difference of speeds |S1 - S2| when moving in the same direction.
Q: Why are train speeds usually converted from km/h to m/s?
Train lengths and platform lengths are typically given in meters and crossing times in seconds. Multiplying km/h by 5/18 converts speed to m/s, keeping units consistent.