Calculate acceleration from speed-time gradients
Use a straight graph's change in speed over its duration for motion in one unchanged direction.
Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.
Before you begin
- Calculate a gradient using two endpoints.
- Use a chosen direction for velocity.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How long is the interval from 7 s to 9 s?
What you will learn
- Calculate straight speed–time gradients.
- Identify constant acceleration and signed deceleration.
Use the axis values and interval
For the speed graph rising from 0 to 8 m/s between zero and four seconds, gradient is (8 − 0)/(4 − 0) = 2 m/s². With motion in one unchanged positive direction, this is acceleration. The positive slope means speeding up; its constant steepness means constant acceleration.
The middle horizontal section has zero gradient and therefore zero acceleration in this straight-line, unchanged-direction model. It still has non-zero speed. Constant acceleration and constant speed are different statements: only the zero-acceleration straight-line case keeps velocity fixed.
Keep the descending gradient's sign
From seven to nine seconds, speed falls from 8 to 0 m/s. Gradient is (0 − 8)/(9 − 7) = −4 m/s². The deceleration magnitude is 4 m/s². Dividing by nine instead of the interval duration two gives the wrong rate.
A speed graph describes magnitude, so it cannot by itself identify acceleration caused only by a turn at constant speed. Relating its gradient to signed acceleration assumes the fixed direction used in these examples. Area gives distance, whereas gradient gives acceleration; the two operations have different units and physical meanings.
Sources for this lesson
Pause and explain
What is acceleration along the graph's horizontal 8 m/s section in straight fixed-direction motion?
Worked example
For fixed positive-direction motion through (0,0), (4,8), (7,8), (9,0), find each acceleration.
Show the worked solution
- First gradient = 8/4 = 2 m/s².
- Middle gradient = 0/3 = 0.
- Final gradient = −8/2 = −4 m/s².
Answer +2, 0 and −4 m/s².
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
Use the diagram's straight speed sections to find accelerations for unchanged positive-direction motion.
Give me a hint
Calculate each gradient from its own endpoints.
Compare my reasoning
- 0–4 s gives +2 m/s².
- 4–7 s gives zero.
- 7–9 s gives −4 m/s².
+2, 0 and −4 m/s².
Look for these in your work
- Use interval differences.
- Keep the descending slope negative.
Speed falls uniformly from 15 to 3 m/s between 2 and 8 s, without changing direction. Find signed acceleration taking motion positive.
Give me a hint
Subtract both endpoint speeds and both endpoint times.
Compare my reasoning
- Δv = 3 − 15 = −12 m/s.
- Δt = 8 − 2 = 6 s.
- a = −2 m/s².
−2 m/s².
Look for these in your work
- Use a six-second interval.
- Identify deceleration.
For the nine-second graph, explain why its area and slope cannot be substituted for one another.
Give me a hint
Compare multiplication of axes with division of their changes.
Compare my reasoning
- Area has unit (m/s) × s = m and gives distance.
- Gradient has unit (m/s)/s = m/s².
- The total distance is 48 m while section accelerations are +2,0,−4 m/s².
Area gives distance; gradient gives an acceleration under the stated direction model.
Look for these in your work
- Check units.
- Keep the physical quantities separate.
Common mistakes
- Subtract the initial time of a section.
- A positive-speed horizontal line has zero acceleration, not zero speed.
- Area and gradient answer different questions.
What can you explain now?
Speed falls from 8 to 0 m/s between 7 and 9 s in a fixed positive direction. What is acceleration?
Compare with the explanation
−4 m/s².
The slope is negative eight m/s over two seconds.
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Make it stick
Annotate a speed graph with one distance area and one acceleration gradient, each with its own unit.
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.