Calculate acceleration from speed-time gradients
Use a straight graph's change in speed over its duration for motion in one unchanged direction.
Section 1.2 · Motion
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028, section 1.2. Build on Core speed and graph skills. Acceleration uses change in velocity per time, not speed divided by time in every situation. A negative acceleration represents deceleration for the positive-direction motion used here. Constant acceleration is not constant speed. Written practice is self-checked; cumulative assessment, supervised practical experience and human subject-teacher review remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
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.
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.
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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.
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.