Speed = distance ÷ time. You can apply it to a whole journey or to just one segment of it, as long as the distance and the time come from the same segment.
This skill is the base of motion and graphs. It appears in short calculation questions and again whenever a graph gradient is asked for.
What does “speed over a segment” mean?
A journey often has parts: speeding up, travelling steadily, stopping. A segment is one part of it, between two chosen times. The speed over that segment uses only the distance covered and the time taken between those two moments.
The word “average” matters. Over a segment the object may change speed, so the result is the average speed for that segment, not the speed at every instant.
How do I calculate it step by step?
- Identify the segment by its start time and end time.
- Find the distance travelled in that segment (final position minus starting position, or the value given).
- Find the time taken: end time minus start time.
- Convert units if needed, so they are consistent (for example, everything in metres and seconds).
- Divide distance by time and write the unit.
Worked example
The table is invented data for a cyclist on a straight road.
| Time (s) | 0 | 20 | 40 | 60 | 80 |
|---|---|---|---|---|---|
| Distance from start (m) | 0 | 100 | 250 | 250 | 400 |
Find the speed from 20 s to 40 s, the speed from 40 s to 60 s, and the average speed for the whole 80 s.
20 s to 40 s: distance = 250 − 100 = 150 m. Time = 40 − 20 = 20 s. Speed = 150 ÷ 20 = 7.5 m/s.
40 s to 60 s: distance = 250 − 250 = 0 m. Speed = 0 ÷ 20 = 0 m/s. The cyclist is stopped.
Whole journey: total distance = 400 m, total time = 80 s. Average speed = 400 ÷ 80 = 5 m/s.
Check for sense: 5 m/s is lower than the 7.5 m/s while moving, which is right because the stop pulls the average down.
The mistake to watch for
A student is told a car travels the first 120 m at 6 m/s and the next 120 m at 12 m/s, and asked for the average speed.
Mistaken answer: (6 + 12) ÷ 2 = 9 m/s.
This treats the two speeds as if the car spent equal time at each.
The car is slower on the first part, so it spends longer there.
Correction: time for the first part = 120 ÷ 6 = 20 s. Time for the second part = 120 ÷ 12 = 10 s. Total distance = 240 m, total time = 30 s. Average speed = 240 ÷ 30 = 8 m/s.
The fix is always the same: add the distances, add the times, then divide.
Check yourself
Try these, then open each answer.
1. A runner covers 360 m in 45 s. What is the speed?
Show answer
Speed = 360 ÷ 45 = 8 m/s.
2. Change 72 km/h into m/s.
Show answer
72 ÷ 3.6 = 20 m/s. Check: 72 000 m ÷ 3600 s = 20 m/s.
3. A train covers 12 km in 10 minutes. Give its average speed in km/h and in m/s.
Show answer
10 minutes = 1/6 hour. Speed = 12 ÷ (1/6) = 72 km/h. In m/s: 12 000 m ÷ 600 s = 20 m/s. The two answers agree, since 72 km/h = 20 m/s.
Where this leads next
Once segment speed is secure, move on to comparing distance-time and speed-time graphs, where the same speed appears as a gradient. When measured values are rounded, the bounds and rounding explainer shows how much a calculated speed could vary.
If you know the formula but still lose marks on units or multi-part journeys, a teacher in online one-to-one Physics tuition can go through your working and find the step that slips.