Time, speed and distance
Distance = speed × time. Convert km/h to m/s by × 5/18, and use relative speed for trains, boats and two people moving together.
Formula sheet
| Situation | Formula |
|---|---|
| Unit change | km/h × 5/18 = m/s; m/s × 18/5 = km/h (72 km/h = 20 m/s) |
| Average speed for a whole trip | Total distance ÷ total time |
| Two equal distances at x and y km/h | Average speed = 2xy ÷ (x + y) |
| Train passing a pole or a person | Time = train length ÷ train speed |
| Train passing a platform or bridge | Time = (train length + platform length) ÷ speed |
| Two trains, opposite directions | Relative speed = sum of speeds; distance = sum of lengths |
| Two trains, same direction | Relative speed = difference of speeds |
| Boat: downstream / upstream | Boat speed + stream speed / boat speed − stream speed |
| Boat and stream from down (d) and up (u) speeds | Boat = (d + u) ÷ 2; stream = (d − u) ÷ 2 |
| Late at speed x, early at speed y | Distance = (xy ÷ (y − x)) × total time difference |
Common traps
- Mixed units: change km/h to m/s (× 5/18) whenever lengths are in metres and time is in seconds.
- Average speed is not the average of the speeds: 60 and 40 km/h over equal distances give 48 km/h, not 50.
- Trains: a pole or a person adds no length; a platform, a bridge or another train adds its own length.
- Minutes and hours: 20 minutes = 1/3 hour, not 0.20 hour.
- Direction: same direction, subtract speeds; opposite directions, add them.
Worked example 1: average speed for a round trip
A dietitian drives to a rural health camp at 60 km/h and returns by the same road at 40 km/h. Find her average speed for the whole trip.
- Standard method: take a convenient one-way distance, the LCM of 60 and 40 = 120 km. Going takes 2 hours and returning takes 3 hours. Total 240 km in 5 hours = 48 km/h.
- Shortcut: equal distances, so average speed = 2xy ÷ (x + y) = 2 × 60 × 40 ÷ 100 = 48 km/h.
- Check: it is below the simple average of 50 km/h because more time is spent at the slower speed.
48 km/h.
Worked example 2: two trains crossing
Two trains, 150 m and 100 m long, run on parallel tracks at 54 km/h and 36 km/h. How long do they take to cross each other when moving (a) in opposite directions and (b) in the same direction?
- Standard method: convert each speed: 54 km/h = 15 m/s and 36 km/h = 10 m/s. Distance to clear = 150 + 100 = 250 m.
- (a) Opposite: relative speed = 15 + 10 = 25 m/s, so time = 250 ÷ 25 = 10 s. (b) Same direction: relative speed = 15 − 10 = 5 m/s, so time = 250 ÷ 5 = 50 s.
- Shortcut: add or subtract in km/h first and convert once: 90 km/h = 25 m/s and 18 km/h = 5 m/s.
- Check: (a) in 10 s the trains cover 150 m + 100 m = 250 m; (b) in 50 s the faster train gains 5 × 50 = 250 m.
(a) 10 seconds; (b) 50 seconds.
Worked example 3: late and early
Walking at 5 km/h, a student reaches college 6 minutes late. Walking at 6 km/h, she is 4 minutes early. How far is the college?
- Standard method: the two times differ by 6 + 4 = 10 minutes = 1/6 hour. So d/5 − d/6 = 1/6, which gives d/30 = 1/6 and d = 5 km.
- Shortcut: distance = (product of speeds ÷ difference of speeds) × time difference = (30 ÷ 1) × 1/6 = 5 km.
- Check: at 5 km/h she takes 60 minutes; at 6 km/h, 50 minutes. The right time is 54 minutes: 6 minutes late and 4 minutes early.
5 km.
Railway and SSC papers favour trains crossing poles, platforms and each other; bank papers add boats and streams and average speed. Placement tests like 'late and early' and meeting-point puzzles, which the relative-speed idea solves quickly.
2 marks: “Convert 90 km/h into metres per second.” 90 × 5/18 = 25 m/s.
5 marks: “A train 240 m long runs at 72 km/h. How long will it take to cross a platform 160 m long?” 72 km/h = 20 m/s; distance = 240 + 160 = 400 m; time = 400 ÷ 20 = 20 seconds.
10 marks: “(a) Find the average speed for equal distances at 60 km/h and 40 km/h. (b) Two trains of 150 m and 100 m at 54 and 36 km/h: find the crossing time in opposite and same directions. (c) At 5 km/h a student is 6 minutes late; at 6 km/h she is 4 minutes early. Find the distance.” (a) 48 km/h. (b) 10 s and 50 s. (c) 5 km. Show unit changes and a check line for each.
D = S × T. km/h to m/s: × 5/18. Equal distances: 2xy/(x + y). Trains: add the lengths; opposite directions add speeds, same direction subtract.