Learning objectives
- Distinguish distance from displacement.
- Calculate speed and acceleration.
- Interpret simple distance-time and velocity-time graphs.
- Use correct SI units.
Distance and displacement
Distance is the total length of the actual path travelled. It is a scalar quantity. Displacement is the shortest straight-line change from initial position to final position and includes direction.
Distance is always non-negative. Displacement can be zero even when distance is not zero, such as after completing one full lap and returning to the starting point.
Speed, velocity and acceleration
Average speed equals total distance divided by total time. Average velocity equals displacement divided by total time. Velocity includes direction.
Acceleration is the rate of change of velocity. In straight-line motion, acceleration in the same direction as velocity increases speed; acceleration in the opposite direction decreases speed. A positive or negative sign depends on the chosen coordinate direction, so negative acceleration does not always mean slowing down.
Reading graphs
The slope of a distance-time graph represents speed. A steeper slope indicates greater speed. A horizontal line means the object is at rest.
The slope of a velocity-time graph represents acceleration. The area under a velocity-time graph represents displacement.
Read motion from a graph
Graphs let you see changes in motion that a single average speed can hide. In the diagram, the object moves, remains at the same distance for a time, then moves again at a greater speed.
Data lab: turn a journey into evidence
A cyclist is 0 m from the start at 0 s, 40 m away at 10 s, still 40 m away at 20 s, and 100 m away at 30 s. Plot these four points and join consecutive points with straight segments.
- 0–10 s: average speed = 4 m/s.
- 10–20 s: speed = 0 m/s because distance does not change.
- 20–30 s: average speed = 6 m/s.
The final section is steeper than the first, so it represents faster motion. This is a direct link between the numerical data and the visual graph.
Worked example: calculate speed and acceleration
A cyclist travels 120 metres in 20 seconds along a straight path. Average speed is total distance divided by elapsed time: 120 ÷ 20 = 6 m/s. This does not prove the cyclist moved at 6 m/s at every instant; some parts of the trip could have been faster or slower.
Now imagine a cyclist's velocity changes steadily from 2 m/s to 8 m/s in 3 seconds in the same direction. Average acceleration is the change in velocity divided by the time taken: (8 − 2) ÷ 3 = 2 m/s². Always include the units and state the direction when a vector is requested.
| Elapsed time | Distance from start |
|---|---|
| 0 s | 0 m |
| 5 s | 20 m |
| 10 s | 40 m |
| 15 s | 60 m |
Plotting the table produces a straight distance–time line. Its slope is 20 m in 5 s, or 4 m/s. A horizontal distance–time line would show no change in distance from the starting point during that interval.
Distance versus displacement
Walk 10 metres east and then 10 metres west. You travelled 20 metres in total, but your final position is the starting point: displacement is zero. Distance records the path length; displacement records the overall change in position.
Practice with numbers
A runner travels 250 m in 50 s. What is the average speed?
Average speed = 250 ÷ 50 = 5 m/s.
A vehicle slows from 18 m/s to 6 m/s in 4 s. What is its average acceleration along its original direction?
Acceleration = (6 − 18) ÷ 4 = −3 m/s². The negative sign indicates acceleration opposite to the chosen positive direction.
Diagnostic: the returning runner
Show a route that goes 30 m east and then 30 m west. Ask for distance first, then displacement. If a student gives 0 m for both, they are treating final position as the whole journey. If they give 60 m for both, they are treating displacement as path length. Ask them to draw an arrow from start to finish before calculating.
Graph check: a flat distance-time segment means distance from the chosen origin is unchanged; it does not mean the graph has “run out of time.”
Motion decisions
Use distance for total path, displacement for directed change in position, speed for distance per time, and velocity for displacement per time. On a distance-time graph, slope is speed; on a velocity-time graph, slope is acceleration and area is displacement.
Practice questions with explanations
Try each question before opening the answer. The explanation shows the reasoning, not only the final response.
Q1. A runner completes one 400 m lap and stops at the start. What are distance and displacement?
Answer: Distance is 400 m; displacement is 0 m.
Explanation: The runner travelled 400 m but finished at the original position.
Q2. Calculate average speed for 150 m travelled in 30 s.
Answer: 5 m/s.
Explanation: Speed = distance ÷ time = 150 ÷ 30.
Q3. What is the SI unit of speed?
Answer: Metre per second (m/s).
Explanation: Distance is measured in metres and time in seconds.
Q4. How does velocity differ from speed?
Answer: Velocity includes direction.
Explanation: Speed is scalar, while velocity is vector.
Q5. A car's velocity changes from 10 m/s to 20 m/s in 5 s. Find acceleration.
Answer: 2 m/s².
Explanation: Acceleration = (20 − 10) ÷ 5.
Q6. What does a horizontal distance-time graph show?
Answer: The object is at rest.
Explanation: Distance does not change as time passes.
Q7. What does the slope of a distance-time graph represent?
Answer: Speed.
Explanation: Slope is change in distance divided by change in time.
Q8. What does the slope of a velocity-time graph represent?
Answer: Acceleration.
Explanation: Slope is change in velocity divided by time.
Q9. Can an object have constant speed but changing velocity?
Answer: Yes.
Explanation: In circular motion the direction changes, so velocity changes.
Q10. What does the area under a velocity-time graph give?
Answer: Displacement.
Explanation: Velocity multiplied by time gives displacement.