Lesson 58 of 100 | Quantitative Aptitude / Speed, Distance and Time / चाल, दूरी और समय
Average speed for unequal times and distances
Learning outcome
Calculate average speed from actual journey totals and recognise when averaging the stated speeds gives an incorrect answer.
Concepts and assumptions
Average speed = total distance travelled ÷ total elapsed time.
It is the constant speed covering that distance in that time, not automatically the arithmetic mean of the individual speeds. On a return journey, include both outward and return distances; returning to the starting point does not make distance zero.
Assume constant speed within each leg, instantaneous speed changes and only stated delays within the departure-to-arrival interval. Convert minutes to hours before combining them with km/h.
For speeds u and v maintained for times t1 and t2:
Average speed = (u × t1 + v × t2) ÷ (t1 + t2).
This is a time-weighted mean: a longer duration gives greater weight. Equal moving times permit (u + v)/2 when there are no stops. Unequal times generally do not.
For two equal positive distances with positive speeds and no stops, let each distance be d. Total time = d/u + d/v. Dividing 2 × d by this time gives:
Average speed = 2 × u × v ÷ (u + v).
The slower leg takes longer, so the arithmetic mean of two different speeds is too high in this case. Do not use this shortcut for unequal distances or journeys containing a stop.
A stop adds time but no distance. Include it when the question asks for the average over the whole journey. Without stops, the average lies between the lowest and highest leg speeds.
Worked examples
Example 1 — Unequal durations. A vehicle travels at 48 km/h for 2 h 30 min and 64 km/h for 1 h 30 min. Times are 2.5 h and 1.5 h. Distances are 48 × 2.5 = 120 km and 64 × 1.5 = 96 km. Average = (120 + 96) ÷ 4 = 54 km/h, not 56 km/h.
Example 2 — Equal distances. A vehicle covers 78 km at 52 km/h and returns over 78 km at 78 km/h. Times are 78/52 = 1.5 h and 78/78 = 1 h. Average = 156 ÷ 2.5 = 62.4 km/h, not the arithmetic mean of 65 km/h.
Example 3 — Unequal distances with a stop. A bus covers 84 km at 42 km/h and 144 km at 48 km/h, stopping for 20 minutes between the legs. Moving times are 2 h and 3 h. Elapsed time = 2 + 3 + 20/60 = 16/3 h. Total distance = 228 km. Average = 228 ÷ (16/3) = 42.75 km/h.
Common mistakes
Do not average speeds without checking their time weights. Do not include a rest in distance. Keep the whole-journey average separate from an average calculated only during movement.
Practice questions
- A vehicle travels for 3 h at 40 km/h and 1 h at 60 km/h, without stops. Find its average speed.
- A rider covers 96 km at 48 km/h and returns over the same distance at 32 km/h, without stops. Find the average.
- A bus covers 72 km at 36 km/h and 126 km at 42 km/h, without stops. Find its average speed.
- A car covers 168 km at 56 km/h whenever moving, with one 30-minute stop. Find its departure-to-arrival average speed.
Worked answers
- Distance = 40 × 3 + 60 × 1 = 180 km. Time = 4 h, so average = 180 ÷ 4 = 45 km/h.
- Times = 96/48 = 2 h and 96/32 = 3 h. Average = 192 ÷ 5 = 38.4 km/h.
- Times = 72/36 = 2 h and 126/42 = 3 h. Total distance = 198 km, so average = 198 ÷ 5 = 39.6 km/h.
- Moving time = 168/56 = 3 h. Elapsed time = 3 + 30/60 = 3.5 h. Average = 168 ÷ 3.5 = 48 km/h.