Beneath trains, boats and two cars on a road lies a single idea — relative speed — sitting on the constant distance = speed × time. Identify the frame of motion and the rest follows.
The base triangle
Distance = Speed × Time (cover any one when given the other two).
Unit conversion
km/h to m/s: multiply by 5/18. m/s to km/h: multiply by 18/5.
Relative speed — the master idea
Two bodies moving in the same direction: subtract their speeds. Moving in opposite directions: add them. Boats use the same rule against the stream.| Scenario | Effective speed |
|---|---|
| Train passes a pole | Length of train / speed |
| Train passes a platform | (Train + platform length) / speed |
| Downstream (with current) | Boat speed + stream speed |
| Upstream (against current) | Boat speed − stream speed |
How to Approach It
- Fix the units before anything else. Convert everything to one system first; km/h becomes m/s by multiplying by 5/18. Mixed units are the single biggest source of wrong answers.
- Decide the frame of motion. Ask whether you have one body or two. Same direction means subtract the speeds; opposite directions means add them; to cross an object you cover the sum of the relevant lengths.
- Spell out the distance covered. A train passing a pole covers only its own length, while passing a platform it covers train plus platform. State exactly what distance is being crossed before dividing.
- Use harmonic mean for round trips. For equal distances at two speeds, the average is 2ab/(a+b), never the plain average. Reach for this whenever a journey is 'there and back'.
Techniques & Methods
- Unit switch ×5/18. km/h → m/s multiply by 5/18; reverse with 18/5. e.g. 72 km/h → 20 m/s.
- Relative speed. Same direction subtract, opposite add; to cross, cover the sum of both lengths. e.g. 30+42 km/h = 20 m/s; 220 m ÷ 20 = 11 s.
- Harmonic average speed. Equal distances at a and b → average = 2ab/(a+b). e.g. 40 and 60 → 48 km/h.
- Boat decomposition. Downstream = b+s, upstream = b−s; so b = (down+up)/2, s = (down−up)/2. e.g. Down 15, up 9 → boat 12, stream 3.
The Edge
Average speed for equal distances at speeds a and b is the harmonic mean 2ab/(a+b), never the plain average. Go at 40 and return at 60 over the same road? Average = 2×40×60/100 = 48, not 50. The plain-average trap is the most common mistake on this topic.Worked example
Convert 72 km/h into metres per second.
- km/h → m/s means ×1000 for km→m and ÷3600 for h→s; combined that is ×5/18.
- Apply the factor: 72 × 5/18.
- Simplify: 72 ÷ 18 = 4, then 4 × 5 = 20.
Answer: 72 km/h = 20 m/s
Worked example
Two trains of length 100 m and 120 m run in opposite directions at 30 km/h and 42 km/h. How long do they take to cross each other?
- Opposite directions → add speeds: 30 + 42 = 72 km/h.
- Convert to m/s: 72 × 5/18 = 20 m/s.
- To fully cross, cover the sum of lengths: 100 + 120 = 220 m.
- Time = 220 ÷ 20 = 11 seconds.
Answer: They cross in 11 seconds
Worked Drills
Worked example
60 km/h expressed in m/s is approximately:
- Multiply by 5/18: 60 × 5/18.
- = 16.67 m/s.
Answer: 16.67 (option c)
Worked example
A 200 m train crosses a 300 m platform in 25 s. Its speed is:
- Distance covered = 200 + 300 = 500 m.
- Speed = 500 / 25 = 20 m/s.
Answer: 20 m/s (option c)
Worked example
A man goes at 40 km/h and returns at 60 km/h over the same route. Average speed:
- Equal distances → harmonic mean 2ab/(a+b).
- = 2×40×60/100 = 48 km/h.
Answer: 48 (option a)
Worked example
A 120 m train passes a man running in the opposite direction at 6 km/h in 6 seconds. The train's speed is:
- Relative speed = 120 / 6 = 20 m/s = 72 km/h.
- Opposite directions, so train = 72 − 6 = 66 km/h.
Answer: 66 km/h (option b)
Worked example
Two stations are 132 km apart. Trains leave towards each other at 20 and 24 km/h. They meet after:
- Closing speed = 20 + 24 = 44 km/h.
- Time = 132 / 44 = 3 h.
Answer: 3 h (option c)
Worked example
A man covers half his journey at 30 km/h and half at 60 km/h. His average speed is:
- Equal distances → harmonic mean 2ab/(a+b).
- = 2×30×60/(30+60) = 3600/90 = 40 km/h.
Answer: 40 km/h (option b)
Worked example
A train crosses a 200 m platform in 30 s and a pole in 10 s. The train's length is:
- Pole: L = 10v. Platform: L + 200 = 30v.
- Subtract: 200 = 20v → v = 10, so L = 10 × 10 = 100 m.
Answer: 100 m (option a)
Worked example
A boat travels 30 km downstream and back in 8 hours. If the stream is 2 km/h, the boat's still-water speed is:
- Let boat speed = b. 30/(b+2) + 30/(b−2) = 8.
- Solve → b = 8 km/h.
Answer: 8 km/h (option b)
Worked example
Walking at three-quarters of his usual speed, a man reaches 20 minutes late. His usual time is:
- Speed ×3/4 → time ×4/3. So (4/3)T = T + 20.
- (1/3)T = 20 → T = 60 min.
Answer: 60 min (option b)
⚠ Watch out
- Average speed is not the average of the two speeds when distances are equal.
- To pass a platform a train must cover its own length plus the platform.
- Convert all units to a single system before calculating.
Takeaways
- Units first. Convert km/h to m/s with ×5/18 before anything else.
- Relative speed rules pairs. Add for opposite directions, subtract for same direction.
- Crossing = sum of lengths. Pole covers train length; platform covers train + platform.
- Round trips use the harmonic mean. 2ab/(a+b), never the plain average.