National 5PhysicsSQARevision NotesDynamicsAccelerationAcceleration

Acceleration (SQA National 5 Physics): Revision Note

Exam code: X857 75

Written by: Katie M

Reviewed by: Caroline Carroll

Updated on 13 October 2025

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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 relationship used to calculate the average acceleration of an object is:

$acceleration = \frac{change in velocity}{time taken}$

$a = \frac{∆ v}{t}$

Where:
- $a$ = acceleration in metres per second squared (m s^-2)
- $∆ v$ = change in velocity in metres per second (m s^-1)
- $t$ = time taken in seconds (s)

Relationship triangle for acceleration, change in velocity and time

Triangle diagram showing velocity change over acceleration and time, with Δv at the top and a and t at the bottom corners. — **To use a relationship triangle, simply cover up the value you wish to calculate and the structure of the relationship will be revealed**

A more detailed explanation of how to use relationship triangles is covered in the revision note on motion relationships
The change in velocity is found by the difference between the initial and final velocity:

$change in velocity = final velocity - initial velocity$

$∆ v = v - u$

Where:
- $v$ = final velocity in metres per second (m s^-1)
- $u$ = initial velocity in metres per second (m s^-1)
Therefore, the relationship for acceleration, or the rate of change of velocity, can be written as:

$a = \frac{v - u}{t}$

Speeding up and slowing down

An object that speeds up is accelerating
An object that slows down is decelerating
Acceleration is a vector quantity, therefore, it 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 (also known as deceleration)

Examples of 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^-1 to 42 m s^-1 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, $u = 50 m s^{- 1}$
Final velocity, $v = 42 m s^{- 1}$

Step 2: Write down the relevant relationship

$∆ v = v - u$

Step 3: Substitute values for final and initial velocity

$∆ v = 42 - 50$

$∆ v = - 8 m s^{- 1}$

Part (b)

Step 1: List the known quantities

Change in velocity, $∆ v = - 8 m s^{- 1}$
- Time taken, $t = 30 s$

Step 2: Write down the relevant relationship

$a = \frac{(v - u)}{t} = \frac{∆ v}{t}$

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

$a = \frac{- 8}{30}$

$a = - 0.27 m s^{- 2}$

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 s^-2

In other words, acceleration measures how much the velocity (in m s^-1) changes every second, m s^-1 s^-1

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