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IGCSE · Cambridge (CIE) · 0652

Calculate acceleration and signed deceleration

Subtract the initial velocity from the final velocity and divide by the elapsed time.

Section P1.2 · Motion

Motion · Cambridge IGCSE Physical Science 0652 · P1.2

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

  • 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.

Sources for this lesson

Pause and explain

With east positive, an eastbound car slows from 18 to 6 m/s in four seconds. What is its acceleration?

Put the idea to work

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
  1. Velocity change = 6 − 18 = −12 m/s.
  2. a = −12/4 = −3 m/s².
  3. Deceleration magnitude is 3 m/s²; acceleration points west.

Answer −3 m/s²; deceleration magnitude 3 m/s².

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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.

1 · Guided

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
  1. Δv = 6 − 18 = −12 m/s.
  2. a = −12/4 = −3 m/s².
  3. 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.
2 · Independent

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
  1. Δv = 10 − 4 = 6 m/s east.
  2. Divide by three seconds.
  3. Mean acceleration = 2 m/s² east.

2 m/s² east.

Look for these in your work

  • Use both endpoints.
  • State direction and units.
3 · Transfer

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
  1. a = (−10 + 4)/3 = −2 m/s².
  2. The velocity magnitude grows from four to ten m/s west.
  3. 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.
Recall without your notes

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.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Calculate one speeding-up and one slowing-down acceleration, stating your positive direction.

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.

Keep learning Physical Science

Course scope and assessment

This page maps the 2025–2026 Cambridge IGCSE Physical Science 0652 syllabus: 38 Chemistry sections and 46 Physics sections. Every section links to original 0620/0625 teaching or one of five new 0652 lessons, with guided, independent and transfer practice and explained answers. Cambridge limits entry to Administrative Zone 3 and identifies November 2026 as the final exam series. The section crosswalk is complete as a first coverage map; a numbered-objective and Core/Supplement audit, independent subject-teacher review and cumulative assessment are still needed. Written practice does not award a Cambridge grade. Online activities cannot demonstrate supervised laboratory competence or complete Paper 5/6.

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