Calculate acceleration and signed deceleration
Subtract the initial velocity from the final velocity and divide by the elapsed time.
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
- Subtract signed values.
- Use consistent velocity and time units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What velocity change takes +18 m/s to +6 m/s?
What you will learn
- Use acceleration = change in velocity/time.
- Interpret negative acceleration for motion in the chosen positive direction.
A velocity change has two endpoints
Acceleration is change in velocity per unit time: a = (v − u)/Δt. A car going east speeds up from 4 to 10 m/s in three seconds. Taking east as positive gives a = (10 − 4)/3 = 2 m/s² east. Dividing the final speed 10 by three would overlook the initial velocity.
The unit m/s² means a velocity change in m/s for each second. An average acceleration calculated from two endpoints does not establish a constant acceleration between them unless that condition is supplied. State the direction or define a signed reference axis before interpreting the sign.
A negative rate can slow positive-direction motion
A vehicle travelling east slows from 18 m/s east to 6 m/s east in four seconds. With east positive, a = (6 − 18)/4 = −3 m/s²: the acceleration is 3 m/s² west and the deceleration magnitude is 3 m/s². Preserve the minus sign if reporting signed acceleration.
Negative acceleration is the syllabus's deceleration case for motion along the positive direction. More generally, slowing means acceleration opposite to velocity. A west-moving object with west negative can speed up under negative acceleration; the sign alone is meaningless without the velocity direction. The practice here states the positive-direction convention explicitly.
Pause and explain
With east positive, an eastbound car slows from 18 to 6 m/s in four seconds. What is its acceleration?
Worked example
An eastbound car slows from 18 to 6 m/s in four seconds. Take east positive. Find acceleration and deceleration magnitude.
Show the worked solution
- Velocity change = 6 − 18 = −12 m/s.
- a = −12/4 = −3 m/s².
- Deceleration magnitude is 3 m/s²; acceleration points west.
Answer −3 m/s²; deceleration magnitude 3 m/s².
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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.
Take east positive. Velocity changes from +18 to +6 m/s in four seconds. Find acceleration.
Give me a hint
Calculate final minus initial before dividing by time.
Compare my reasoning
- Δv = 6 − 18 = −12 m/s.
- a = −12/4 = −3 m/s².
- This is westward acceleration, slowing the eastbound car.
−3 m/s², or 3 m/s² west.
Look for these in your work
- Keep the signed change.
- Interpret it relative to motion.
An eastbound vehicle increases velocity from 4 to 10 m/s in three seconds. Find mean acceleration.
Give me a hint
Use the velocity increase rather than the final speed alone.
Compare my reasoning
- Δv = 10 − 4 = 6 m/s east.
- Divide by three seconds.
- Mean acceleration = 2 m/s² east.
2 m/s² east.
Look for these in your work
- Use both endpoints.
- State direction and units.
With east positive, velocity changes from −4 to −10 m/s in three seconds. Is the object slowing? Find acceleration.
Give me a hint
Compare magnitudes and directions, not the acceleration sign alone.
Compare my reasoning
- a = (−10 + 4)/3 = −2 m/s².
- The velocity magnitude grows from four to ten m/s west.
- It speeds up westward; negative acceleration does not mean slowing in this signed example.
−2 m/s²; it speeds up westward.
Look for these in your work
- Retain signed velocities.
- Distinguish sign from slowing.
Common mistakes
- Use final minus initial velocity.
- An endpoint average does not by itself prove a constant rate.
- A minus sign needs a chosen direction.
What can you explain now?
Velocity rises from 4 to 10 m/s in three seconds. What is mean acceleration?
Compare with the explanation
2 m/s².
The change is six m/s over three seconds.
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Make it stick
Calculate one speeding-up and one slowing-down acceleration, stating your positive direction.
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