Core relation
Distance = Speed x Time. Convert km/h to m/s by multiplying by 5/18; m/s to km/h by 18/5. 36 km/h = 10 m/s, 54 km/h = 15 m/s, 72 km/h = 20 m/s, 90 km/h = 25 m/s. Target: 45 seconds.
Key formulas
| Situation | Formula |
|---|---|
| Average speed, equal distances at x and y | 2xy/(x + y) |
| Average speed, three equal distances | 3xyz/(xy + yz + zx) |
| Relative speed, same direction | x - y |
| Relative speed, opposite directions | x + y |
| Train crossing a pole/man standing | length of train / speed |
| Train crossing platform/bridge | (train + platform)/speed |
| Two trains crossing | (L1 + L2)/relative speed |
| Late/early problem | d = (product of speeds/difference of speeds) x time difference |
| After meeting, times a and b to destinations | speed ratio = sqrt(b) : sqrt(a); meeting time = sqrt(ab) |
Constant-distance idea
For a fixed distance, speed and time are inversely proportional. If speed becomes 4/5, time becomes 5/4 - an increase of 1/4 of the usual time. If this extra time is 15 minutes, the usual time is 60 minutes.
Worked example (train)
A 180 m train at 54 km/h crosses a 120 m platform. Time?
Speed = 15 m/s. Distance = 300 m. Time = 20 seconds.
Worked example 2 (two trains)
Trains of 100 m and 150 m run in opposite directions at 60 km/h and 40 km/h. Relative speed = 100 km/h = 250/9 m/s. Time = 250 / (250/9) = 9 seconds.
Worked example 3 (average speed)
Going at 40 km/h and returning at 60 km/h: average = 2 x 40 x 60/100 = 48 km/h, not 50.
Chasing and meeting
- Thief and policeman: time to catch = gap / (difference of speeds).
- Two people start towards each other: time to meet = distance / (sum of speeds).
- Circular track, same direction: they meet first after track length / (difference of speeds).
Common traps
- Simple average of speeds for equal distances.
- Forgetting to add the platform length.
- Using same-direction relative speed for opposite directions.
- Mixing minutes into hour calculations (6 minutes = 1/10 hour).
Worked example 4 (late and early)
At 5 km/h a man is 6 minutes late; at 6 km/h he is 4 minutes early. d = (5 x 6)/(6 - 5) x 10/60 = 30 x 1/6 = 5 km.
Worked example 5 (fraction of usual speed)
Walking at 3/4 of his usual speed, a man is 20 minutes late. Time becomes 4/3 of usual, so the extra 1/3 of usual time = 20 minutes, and the usual time = 60 minutes.