Distinguish constant and changing acceleration
Read straight and curved speed-time shapes and use equal-time velocity changes to justify a trend.
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
- Link a speed–time gradient to acceleration for fixed-direction motion.
- Compare changes over equal durations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What gradient describes a straight speed–time segment?
What you will learn
- Recognise constant acceleration from a straight speed–time segment.
- Interpret a curved graph using its changing gradient.
Equal changes per second
For fixed-direction motion with speeds 0, 2, 4, 6, 8 m/s at times 0,1,2,3,4 s, each one-second change is 2 m/s. A straight speed–time line through these data is consistent with constant acceleration 2 m/s². The speeds themselves are not constant.
Speeds 0, 1, 4, 9, 16 m/s at the same times instead change by 1,3,5,7 m/s. The interval mean accelerations increase. A smooth illustrative speed law v = t² has a steepening curve and increasing local gradient. A sparse data table establishes interval averages, not a unique continuous curve between measurements.
A curved gradient is local
On a smooth speed–time curve, a tangent at an instant estimates its gradient and hence the instantaneous acceleration for unchanged positive-direction motion. A chord between two separated times gives mean acceleration over that interval. Do not mistake a whole-interval chord for every local acceleration.
A rising curve that flattens has decreasing positive acceleration; a curve that steepens has increasing positive acceleration. A falling line indicates slowing in this positive-direction model, with a negative slope. State how the gradient changes rather than saying only that the graph is curved. Drawing and reading tangents has a precision limit.
Pause and explain
A speed–time curve rises and becomes progressively steeper. What happens to acceleration?
Worked example
Speeds at one-second intervals are 0,1,4,9,16 m/s. Find interval mean accelerations and describe their trend.
Show the worked solution
- Velocity changes are 1,3,5,7 m/s.
- Each interval lasts one second.
- Mean accelerations are 1,3,5,7 m/s², increasing across intervals.
Answer 1,3,5,7 m/s²; acceleration is changing.
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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.
Speeds at 0,1,2,3,4 s are 0,2,4,6,8 m/s. Find interval accelerations.
Give me a hint
Each interval lasts one second.
Compare my reasoning
- Successive changes are all 2 m/s.
- Divide each by one second.
- Each mean acceleration is 2 m/s².
2 m/s² in every supplied interval.
Look for these in your work
- Compare equal-time changes.
- Distinguish average-data evidence from exact between-point motion.
Speeds at the same times are 0,1,4,9,16 m/s. Find interval accelerations and the 0–4 s mean.
Give me a hint
Calculate successive changes, then the full-interval change separately.
Compare my reasoning
- Interval accelerations are 1,3,5,7 m/s².
- Overall change is 16 m/s over four seconds.
- Overall mean acceleration is 4 m/s².
1,3,5,7 m/s²; overall mean 4 m/s².
Look for these in your work
- Use time differences.
- Separate local intervals from the whole mean.
For the illustrative smooth curve v = t², a tangent at t = 2 s passes through (1 s,0 m/s) and (3 s,8 m/s). Estimate acceleration there.
Give me a hint
Use two widely separated points on the stated tangent, not two arbitrary curve points.
Compare my reasoning
- Tangent speed rise = 8 m/s.
- Tangent time run = 2 s.
- Gradient = 4 m/s² at that instant; tangent points need not lie on the original curve.
4 m/s².
Look for these in your work
- Use the tangent line.
- Identify an instantaneous estimate.
Common mistakes
- Constant acceleration is not constant speed.
- A chord gives a mean rather than a local tangent rate.
- Data points alone do not determine all intervening acceleration.
What can you explain now?
What shape shows constant non-zero acceleration on a fixed-direction speed–time graph?
Compare with the explanation
A straight sloping line.
Its gradient is constant and non-zero.
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Make it stick
Sketch a constant-acceleration line and a changing-acceleration curve, then explain tangent and chord gradients.
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