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O LevelPhysicsCambridge (CIE)Revision Notes1. Motion, Forces & Energy1.2 MotionAcceleration

Acceleration (Cambridge (CIE) O Level Physics): Revision Note

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Acceleration

  • Acceleration is defined as the rate of change of velocity

    • In other words, it describes how much an object's velocity changes every second

  • The equation below is used to calculate the average acceleration of an object:

 a c c e l e r a t i o n space equals space fraction numerator c h a n g e space i n space v e l o c i t y space over denominator c h a n g e space i n space t i m e end fraction

a space equals space fraction numerator increment v over denominator increment t end fraction

  • Where:

    • a = acceleration in metres per second squared (m/s2)

    • Δv = change in velocity in metres per second (m/s)

    • Δt = time taken in seconds (s)

  • The change in velocity is found by the difference between the initial and final velocity, as written below:

change in velocity = final velocity − initial velocity

increment v space equals space v space minus space u

  • Where:

    • v = final velocity in metres per second (m/s)

    • u = initial velocity in metres per second (m/s)

  • The equation for acceleration can be rearranged with the help of a formula triangle as shown:

Acceleration equation triangle

1-2-2-acceleration-triangle-cie-igcse-23-rn

Equation triangle for the change in velocity, acceleration, and change in time

Speeding Up & Slowing Down

  • An object that speeds up is accelerating

  • An object that slows down is decelerating

  • The acceleration of an object can be positive or negative, depending on whether the object is speeding up or slowing down

    • If an object is speeding up, its acceleration is positive

    • If an object is slowing down, its acceleration is negative (sometimes called deceleration)

Acceleration and deceleration

Acceleration Examples, downloadable IGCSE & GCSE Physics revision notes

A rocket speeding up (accelerating) and a car slowing down (decelerating)

Worked Example

A Japanese bullet train decelerates at a constant rate in a straight line. The velocity of the train decreases from 50 m/s to 42 m/s in 30 seconds.

(a) Calculate the change in velocity of the train.

(b) Calculate the deceleration of the train, and explain how your answer shows the train is slowing down.

Answer:

Part (a)

Step 1: List the known quantities

  • Initial velocity = 50 m/s

  • Final velocity = 42 m/s

Step 2: Write the relevant equation

change in velocity = final velocity − initial velocity

Step 3: Substitute values for final and initial velocity

change in velocity = 42 − 50 = −8 m/s

Part (b)

Step 1: List the known quantities

  • Change in velocity, Δv = − 8 m/s

  • Time taken, t = 30 s

Step 2: Write the relevant equation

 a space equals space fraction numerator increment v over denominator increment t end fraction

Step 3: Substitute the values for change in velocity and time

a = −8 ÷ 30 = −0.27 m/s

Step 4: Interpret the value for deceleration

  • The answer is negative, which indicates the train is slowing down

Examiner Tips and Tricks

Remember the units for acceleration are metres per second squared, m/s2. In other words, acceleration measures how much the velocity (in m/s) changes every second, m/s/s.

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    Ashika

    Author: Ashika

    Expertise: Physics Project Lead

    Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.