60-second revision
- A train passing a pole or a standing man covers its own length.
- A train crossing a platform, bridge or tunnel covers its length + the platform length.
- Two trains: add speeds if opposite, subtract if same direction; distance = sum of both lengths.
- Downstream = u + v, upstream = u − v.
- Still water u = (D + U)/2; stream v = (D − U)/2.
- Upstream takes k times as long as downstream: u:v = (k + 1):(k − 1).
- Two platforms: speed = difference in platform lengths ÷ difference in times.
Trains: pick the right distance
A train is not a point. Whatever it crosses, the distance is "front of the train enters" to "back of the train leaves". Decide the distance and the relative speed, convert to m/s, then divide.
| Train crosses | Distance covered | Speed to use |
|---|---|---|
| A pole, signal, tree or standing man | Train length | Train speed |
| A platform, bridge or tunnel | Train length + platform (bridge, tunnel) length | Train speed |
| A man walking towards it | Train length | Train speed + man's speed |
| A man walking the same way | Train length | Train speed − man's speed |
| Another train, opposite direction | Sum of both lengths | Sum of speeds |
| Another train, same direction | Sum of both lengths | Difference of speeds |
Worked example: a 150 m train at 45 km/h (12.5 m/s) crosses a 350 m bridge. Distance 500 m, time = 500/12.5 = 40 s.
Worked example: trains of 120 m and 130 m run the same way at 50 and 40 km/h. Relative speed = 10 km/h = 25/9 m/s. Time to pass = 250 × 9/25 = 90 s.
Trains meeting from two stations
If two trains start from stations A and B towards each other and, after meeting, take a and b hours to reach their destinations, then (speed of first) : (speed of second) = sqrt(b) : sqrt(a). Example: after meeting they take 9 h and 4 h; speed ratio = 2:3.
Boats and streams
Let u = speed of the boat in still water and v = speed of the stream. With the current, speeds add; against it, they subtract.
| Quantity | Formula |
|---|---|
| Downstream speed D | u + v |
| Upstream speed U | u − v |
| Speed in still water | (D + U)/2 |
| Speed of stream | (D − U)/2 |
| Round trip of d km | d/(u + v) + d/(u − v) |
| Average speed of round trip | (u² − v²)/u |
Worked example: a boat goes 60 km downstream in 4 hours and 33 km upstream in 3 hours. D = 15, U = 11, so u = 13 km/h and v = 2 km/h.
Worked example: a boat with u = 8 km/h on a river with v = 2 km/h goes 30 km and returns. Time = 30/10 + 30/6 = 3 + 5 = 8 hours.
Worked example: a swimmer covers 6 km downstream in 1 hour and takes 3 hours to swim back. D = 6 and U = 2, so she swims at 4 km/h in still water and the current flows at 2 km/h.
Worked example: a train 200 m long at 68 km/h passes a man walking towards it at 4 km/h. Relative speed = 72 km/h = 20 m/s, so it passes him in 10 seconds. Had he walked the same way, the relative speed would be 64 km/h and it would take 11.25 seconds.
Method in four steps
- Draw the situation: who moves, which way, and what is being crossed.
- Write the distance (one length, two lengths, or length + platform).
- Write the relative speed and convert it to m/s if lengths are in metres.
- Divide, then re-read what is asked (time, speed in km/h, or a length).
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