Time, distance and work
Distance is speed × time, and work is shared by adding daily rates. Convert units first and use relative speed for trains and boats.
Worked example: average speed
A student cycles to college at 12 km/h and returns by the same road at 18 km/h. Find the average speed for the round trip.
- Equal distances at different speeds: use 2xy ÷ (x + y).
- 2 × 12 × 18 ÷ (12 + 18) = 432 ÷ 30 = 14.4 km/h.
- Check with 36 km each way: 3 h + 2 h = 5 h for 72 km, and 72 ÷ 5 = 14.4 km/h.
14.4 km/h, not 15. Shortcut: the average speed over equal distances is the harmonic mean, which is always below the simple average.
Worked example: trains and relative speed
A 200 m train runs at 72 km/h. How long does it take to cross a 160 m platform? How long does it take to pass a man running at 12 km/h in the same direction?
- Convert: 72 km/h × 5/18 = 20 m/s.
- Platform: distance = 200 + 160 = 360 m; time = 360 ÷ 20 = 18 s.
- Man, same direction: relative speed = 72 − 12 = 60 km/h = 60 × 5/18 = 50/3 m/s.
- Distance = the train's length only, 200 m; time = 200 ÷ (50/3) = 600 ÷ 50 = 12 s.
18 s for the platform; 12 s to pass the man. Shortcut: a pole or a person adds no length; a platform or another train adds its own length.
Worked example: work done together
A can finish a job in 12 days and B in 18 days. How long will they take together? If A leaves after 4 days, how long will B take to finish the rest?
- Take total work = LCM(12, 18) = 36 units.
- A does 36 ÷ 12 = 3 units a day and B does 36 ÷ 18 = 2 units a day; together, 5 units a day.
- Together: 36 ÷ 5 = 7.2 days.
- In 4 days together they finish 4 × 5 = 20 units, leaving 16 units.
- B alone: 16 ÷ 2 = 8 more days.
7.2 days together; B needs 8 more days. Shortcut: take the LCM as the total work so every daily rate is a whole number.
Formula sheet
| Situation | Formula |
|---|---|
| Basic | Distance = speed × time |
| Units | km/h × 5/18 = m/s; m/s × 18/5 = km/h |
| Average speed, equal distances | 2xy ÷ (x + y) |
| Relative speed | Opposite directions: add; same direction: subtract |
| Train crossing | Time = (train length + object length) ÷ relative speed |
| Boats and streams | Downstream = b + s; upstream = b − s; b = (down + up) ÷ 2 |
| Work | If A finishes in a days, A does 1/a of the work per day |
| Workers and days | M₁D₁ = M₂D₂ (with hours: M₁D₁H₁ = M₂D₂H₂) |
NET asks unit conversion, round-trip average speed, train crossings and two- or three-worker problems. Check the trap first: equal distances need the harmonic mean, not the simple average.
Typical MCQ: “A car covers 150 km in 2.5 hours. What is its speed in m/s?” 150 ÷ 2.5 = 60 km/h, and 60 × 5/18 = 50/3 ≈ 16.67 m/s.
Typical MCQ: “12 workers finish a job in 15 days. How many workers are needed to finish it in 10 days?” 18. By M₁D₁ = M₂D₂, 12 × 15 = M₂ × 10, so M₂ = 18.
5/18 shrinks (km/h to m/s), 18/5 grows (m/s to km/h). Round-trip average = 2xy/(x + y), never (x + y)/2. For work, LCM = total work.