Describe motion with distance, displacement, speed and velocity
Separate path length from change in position, and calculate average speed and velocity while preserving direction and units.
AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 2. The falling-object activity is an online description and calculation practice; it does not provide supervised laboratory competence. Full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use SI units and signed values, and recall that velocity includes direction.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which quantity includes the direction of motion?
What you will learn
- Define and use distance and displacement
- Define and use speed and velocity
- Calculate average acceleration from a change in velocity when needed
Position, path and direction
Distance is the total length of the path travelled. Displacement is the change in position from the starting point and includes direction. A person can travel a non-zero distance and finish with zero displacement after returning to the start.
Choose an origin and a positive direction before using signed coordinates. Displacement is final position minus initial position. In one dimension, a negative result means the change is opposite to the chosen positive direction; it is not a negative distance.
Average quantities use the whole interval
Average speed is total distance divided by elapsed time. Average velocity is displacement divided by elapsed time. Because distance and displacement can differ, average speed and the magnitude of average velocity need not be equal.
Velocity is a vector; speed is its magnitude. Average acceleration is change in velocity divided by elapsed time. Keep initial and final velocity signs consistent with the same direction axis.
Pause and explain
A runner completes one lap and returns to the start. Which statement is correct?
Worked example
A student walks 100 m east, then 60 m west in 40 s. Find distance, displacement, average speed and average velocity.
Show the worked solution
- Distance is the full path: 100 + 60 = 160 m.
- Take east as positive. Displacement is +100 − 60 = +40 m, or 40 m east.
- Average speed is 160/40 = 4.0 m s⁻¹; average velocity is 40/40 = 1.0 m s⁻¹ east.
Answer Distance 160 m; displacement 40 m east; average speed 4.0 m s⁻¹; average velocity 1.0 m s⁻¹ east.
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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.
A lift moves 12 m up and then 5 m down in 17 s. Find its distance, displacement, average speed and average velocity.
Give me a hint
Use upward as positive; calculate path and net change separately.
Compare my reasoning
- Distance = 12 + 5 = 17 m.
- Displacement = +12 − 5 = +7 m.
- Divide distance and displacement separately by 17 s.
Distance 17 m; displacement 7 m up; average speed 1.0 m s⁻¹; average velocity about 0.41 m s⁻¹ up.
Look for these in your work
- Path length and displacement distinguished
- Both averages use the full elapsed time
- Direction retained for velocity
A car travels 240 m east in 20 s, then 160 m west in 20 s. Find the average speed and average velocity for the whole trip.
Give me a hint
The net displacement uses signed motion; the distance adds both parts.
Compare my reasoning
- Total distance = 240 + 160 = 400 m.
- With east positive, displacement = 240 − 160 = 80 m east.
- Total time = 40 s; calculate both averages.
Average speed = 10 m s⁻¹; average velocity = 2.0 m s⁻¹ east.
Look for these in your work
- Total path and net change calculated separately
- Elapsed times added
- Average quantities and direction correct
A particle's velocity changes from 6.0 m s⁻¹ east to 2.0 m s⁻¹ west in 4.0 s. Taking east as positive, find its average acceleration.
Give me a hint
Represent westward final velocity with a negative sign.
Compare my reasoning
- Initial velocity u = +6.0 m s⁻¹.
- Final velocity v = −2.0 m s⁻¹, so change = −8.0 m s⁻¹.
- Average acceleration = (v − u)/t = −8.0/4.0.
−2.0 m s⁻², meaning 2.0 m s⁻² west.
Look for these in your work
- Direction signs consistent
- Velocity change calculated
- Acceleration direction interpreted
Common mistakes
- Using total distance to calculate average velocity
- Reporting displacement without its direction or sign convention
- Assuming average speed always equals the magnitude of average velocity
What can you explain now?
Which quantity includes the direction of motion?
Compare with the explanation
Velocity
Velocity is a vector and includes direction.
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Make it stick
How do distance and displacement differ, and why can average speed differ from average velocity?
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.