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IGCSE · Cambridge (CIE) · 0653

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 Combined Science 0653 · P1.2

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

  • 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 Science — Combined

Course scope and assessment

This pathway maps all 121 detailed subject-content sections in the published 0653 syllabus for exams in 2028 and 2029. It adapts original lessons from the related single-science pathways and keeps Combined Science's narrower scope: Biology B1–B16, Chemistry C1–C12 and Physics P1–P5. A 0653-specific objective-by-objective review, Core/Supplement audit, human subject-teacher review, supervised laboratory experience and cumulative assessment remain in progress. Written practice is self-checked and does not predict an official grade. The separate 2025–2027 syllabus has not been reconciled here; choose the 2028–2029 syllabus only.

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