Distinguish constant and changing acceleration
Read straight and curved speed-time shapes and use equal-time velocity changes to justify a trend.
Section P1.2 · Motion
Cross-referenced to the 2028–2029 0653 section using original teaching from the related single-science courses. A 0653-specific objective-by-objective and Core/Supplement review is in progress. Written practice is self-checked and does not demonstrate supervised laboratory competence or award an official Cambridge grade.
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.
Sources for this lesson
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.
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.