Find whole-journey average speed including stops
Add all travelled distances and all elapsed times rather than averaging speed values blindly.
Section 1.2 · Motion
Exam scope and assessment notes
Core and Extended · Physics 0625, 2026–2028, section 1.2. Core includes speed, velocity direction, distance–time and speed–time interpretation, straight distance–time gradients, speed–time areas and near-Earth free-fall acceleration. Numerical acceleration and the resistance/terminal-velocity explanation are Supplement requirements, taught in separately labelled Extended lessons. Written practice is self-checked; cumulative assessment, supervised practical experience and human subject-teacher review remain pending.
Before you begin
- Add journey segments.
- Calculate a distance-per-time rate.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What distance is added during a ten-second stop?
What you will learn
- Calculate total distance divided by total time.
- Explain why an arithmetic mean of speeds can be wrong.
The whole journey sets the denominator
Average speed is total distance travelled divided by total time taken. For 100 m in 20 s, a ten-second stop, then 100 m in 40 s, the total is 200 m in 70 s. The whole-journey average is 200/70 ≈ 2.86 m/s. The stop contributes time but no additional distance.
Use only the interval specified in the question. If it asks for the average while moving, exclude the stopped interval and state that restriction. If it asks from departure to arrival, include it. Different denominators answer different questions; neither can be chosen merely to produce a larger speed.
Speeds need time weighting
Averaging two speeds as (v₁ + v₂)/2 is valid for equal time spent at each speed. It is generally wrong for equal distances, because the slower leg takes longer. For 60 km at 30 km/h and another 60 km at 60 km/h, total time is 2 + 1 = 3 h and average speed is 120/3 = 40 km/h, not 45 km/h.
Returning to the start gives zero displacement, but the travelled distance is still the sum of the outward and return paths. Average speed is therefore positive for a non-zero journey. The directed average velocity can be zero; do not replace the speed numerator with net displacement.
Pause and explain
Equal 60 km legs take 2 h and 1 h. Which calculation gives average speed?
Worked example
A walker covers 100 m in 20 s, rests for 10 s, then covers 100 m in 40 s. Find whole-journey average speed.
Show the worked solution
- Total distance = 200 m.
- Total elapsed time = 20 + 10 + 40 = 70 s.
- Average speed = 200/70 ≈ 2.86 m/s.
Answer Approximately 2.86 m/s.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find average speed for 100 m in 20 s, a 10 s stop and 100 m in 40 s.
Give me a hint
Include all three intervals but only moving distances.
Compare my reasoning
- Total distance is 200 m.
- Total time is 70 s.
- The rate is approximately 2.86 m/s.
Approximately 2.86 m/s.
Look for these in your work
- Include the stopped time.
- Use total distance.
A car travels 60 km at 30 km/h and 60 km at 60 km/h. Find average speed.
Give me a hint
Calculate the time of each leg before combining them.
Compare my reasoning
- First time = 60/30 = 2 h.
- Second time = 60/60 = 1 h.
- 120 km / 3 h = 40 km/h.
40 km/h.
Look for these in your work
- Find both durations.
- Avoid a blind arithmetic speed mean.
A runner travels 400 m around a loop in 100 s and returns to the start. Find average speed and explain why it is not zero.
Give me a hint
The numerator is path length rather than net position change.
Compare my reasoning
- Total distance is 400 m.
- 400/100 = 4 m/s.
- Net displacement is zero, but that is not the average-speed numerator.
4 m/s; the full path still counts.
Look for these in your work
- Use travelled distance.
- Separate it from displacement.
Common mistakes
- Include a stop in departure-to-arrival elapsed time.
- Equal distance does not imply equal time.
- Distance is not net displacement.
What can you explain now?
For equal 60 km legs at 30 and 60 km/h, what is average speed?
Compare with the explanation
40 km/h.
The legs take two hours and one hour, so divide 120 km by three hours.
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Make it stick
Build a two-leg journey with a stop and calculate both departure-to-arrival and moving-only averages.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.