Average speed is total distance ÷ total time, not (speed 1 + speed 2) ÷ 2. The two formulas only agree when the time spent at each speed is identical, which almost never happens in an exam question. Whenever a journey has two legs of equal distance (not equal time), you must add up the actual distance and the actual time separately, then divide, or you will get the wrong answer.
It feels natural to see two speeds, say 20 km/h and 30 km/h, and average them to 25 km/h. This shortcut only works if the object travels at each speed for the same amount of time. In almost every O-Level average speed question, the object instead travels the same distance at each speed, which means it spends longer at the slower speed and less time at the faster one. That imbalance pulls the true average speed below the simple average of the two speeds.
The safe rule that always works, no matter how the journey is split, is to go back to the definition: average speed equals total distance divided by total time.
Step 1
Find the distance for each leg, add for total distance
Step 2
Find the time for each leg (include any rest stops), add for total time
Step 3
Average speed = total distance ÷ total time
Step 4
Never average the two speeds directly
A cyclist rides 60 km to a town at a steady 20 km/h, then rides the same 60 km back home at a steady 30 km/h. Find the average speed for the whole journey.
Solution
60 km + 60 km = 120 km.60 ÷ 20 = 3 h. Time for the return leg: 60 ÷ 30 = 2 h.3 h + 2 h = 5 h.120 ÷ 5 = 24 km/h.Notice: the simple average of 20 and 30 is 25 km/h, but the correct answer is 24 km/h. The cyclist spends more time (3 h) riding at the slower speed than at the faster one (2 h), so the true average is pulled towards the slower speed.
A driver travels 90 km from town A to town B at 45 km/h, stops for a 30-minute break, then drives a further 60 km to town C at 40 km/h. Find the average speed for the whole journey from A to C.
Solution
90 km + 60 km = 150 km.90 ÷ 45 = 2 h.30 min = 0.5 h. This counts towards the total time, because the driver is not moving but time is still passing.60 ÷ 40 = 1.5 h.2 h + 0.5 h + 1.5 h = 4 h.150 ÷ 4 = 37.5 km/h.The step students get wrong
Leaving out the rest stop. Average speed uses the total time for the whole journey, including any time spent stationary, not just the time spent actually moving. If the 30-minute break above were left out, the total time would wrongly become 3.5 h and the average speed would come out as 150 ÷ 3.5 ≈ 42.9 km/h, which overstates how fast the journey really went. Always check the question for words like "rest", "stopped", or "break" and add that time in.
Quick check: when the object is moving for the whole journey (no rest stops), the average speed must always land between the slowest and fastest speeds used. In worked example 1, 24 km/h sits between 20 and 30. This check does not apply once a rest stop is added, as in worked example 2: the idle time pulls the average speed of 37.5 km/h below both moving speeds of 40 and 45 km/h. If your answer falls outside the expected range and there is no rest stop in the question, you have made an arithmetic slip.
Is average speed ever equal to the average of the two speeds?
Only when the object travels at each speed for exactly the same length of time, not the same distance. Most O-Level questions describe equal-distance legs, so this special case rarely applies. When in doubt, always calculate total distance divided by total time instead of guessing.
Does average speed use displacement or total distance travelled?
Total distance travelled, not displacement. If a car drives 50 km out and 50 km back to its starting point, its displacement is zero, but it has still covered 100 km, and that 100 km is what goes into the average speed formula.
How do I handle time given in minutes when the speed is in km/h?
Convert minutes to hours before adding to the other times, since the speed unit sets the time unit. Divide the number of minutes by 60, so 30 minutes becomes 0.5 h and 15 minutes becomes 0.25 h.
Where can I see this idea plotted out visually?
A distance-time graph makes the unequal time split visible directly, since the slower leg produces a longer, flatter section of the graph. Our guide to reading distance-time and speed-time graphs covers how to pull speeds and times off each type.
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