Calculate acceleration and signed deceleration
Subtract the initial velocity from the final velocity and divide by the elapsed time.
Section 1.2 · Motion
Exam scope and assessment notes
Extended only (Supplement) · Physics 0625, 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.
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.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →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.
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.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
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.