AQA GCSE Combined Science Foundation (8464), Foundation tier · Physics › Forces › Forces and motion › Describing motion along a line
Practise Acceleration. 5 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Acceleration is the rate of change of velocity. You need to use a = Δv ÷ t and v2 − u2 = 2 a s, and find acceleration from the gradient of a velocity–time graph. You also need free fall and terminal velocity. This subtopic produces many calculation questions.
Key facts
a = Δv ÷ t (acceleration = change in velocity ÷ time taken)
v2 − u2 = 2 a s (on the equations sheet)
Unit of acceleration: m/s2
Deceleration = negative acceleration
Free-fall acceleration near the Earth's surface ≈ 9.8 m/s2
v–t graph: gradient = acceleration
Terminal velocity: drag = weight, resultant force = 0
Notes
Acceleration
Acceleration is the rate of change of velocity: a = Δv ÷ t
a in m/s2, Δv (change in velocity = final velocity − initial velocity) in m/s, t in s.
An object that slows down is decelerating. Its acceleration is negative.
Near the Earth's surface, any object falling freely under gravity has an acceleration of about 9.8 m/s2.
Everyday example: a car going from 0 to 30 m/s in about 10 s has an acceleration of about 3 m/s2.
Uniform acceleration: v2 − u2 = 2 a s
(final velocity)2 − (initial velocity)2 = 2 × acceleration × distance. This equation is on the equations sheet.
Use it when the acceleration is uniform and the question gives no time. u = initial velocity, v = final velocity, s = distance.
Example: a ball dropped from rest falls 5.0 m. v2 = 0 + 2 × 9.8 × 5.0 = 98, so v = 9.9 m/s.
Velocity–time graphs
diagram
Velocity is on the y-axis and time is on the x-axis.
Horizontal line: constant velocity. Straight line sloping up: constant acceleration. Straight line sloping down: constant deceleration.
Gradient = acceleration. A steeper line means a greater acceleration; a curved line means the acceleration is changing.
On a velocity–time graph the gradient is the acceleration.
Falling and terminal velocity
diagram
An object falling through a fluid (air or a liquid) initially accelerates due to the force of gravity.
As its speed increases, the drag (air resistance) on it increases, so the resultant force decreases.
Eventually the resultant force is zero and the object moves at a steady speed: its terminal velocity.
As the skydiver speeds up, air resistance grows until it balances the weight.
How to answer each type of question
Calculate acceleration
2 to 3 marksGrade 4
Find the change in velocity: final − initial.
Divide by the time taken.
Give the unit m/s2.
Example. A train speeds up from 8.0 m/s to 20.0 m/s in 40 s. Calculate its acceleration.
Show the model answerHide the model answer
Δv = 20.0 − 8.0 = 12.0 m/s (1) a = 12.0 ÷ 40 (1) a = 0.30 m/s2 (1)
Don’t lose marks
Using the final velocity instead of the change in velocity in a = Δv ÷ t.
Forgetting to square u and v, or forgetting to take the square root at the end, in v2 − u2 = 2 a s.
Reading a velocity–time graph as if it were a distance–time graph: a horizontal line on a v–t graph means constant velocity, not stationary.
Saying there are no forces on an object at terminal velocity. The forces are balanced: drag = weight.
Giving acceleration in m/s instead of m/s2.
More tips
Memory tricks
Velocity–time graph: slope = acceleration.
No time in the question? Use v2 − u2 = 2 a s. Time given? Use a = Δv ÷ t.
'Starts from rest' means u = 0; 'comes to rest' means v = 0.
Sense check: road vehicles accelerate at a few m/s2, not hundreds.
Terminal velocity: forces balanced, speed steady.
Exam technique
a = Δv ÷ t must be recalled; v2 − u2 = 2 a s is on the equations sheet.
For graph questions, draw on the graph and write down the values you read.
In terminal velocity explanations, describe what happens to the air resistance, then the resultant force, then the acceleration, in that order.
For deceleration in v2 − u2 = 2 a s, use a negative value of a, or work with sizes and say 'deceleration'.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade 4
Calculate acceleration using a = Δv ÷ tFrom 4 m/s to 16 m/s in 3 s: a = 12 ÷ 3 = 4 m/s2.
Grade 4
Describe motion from a velocity–time graphHorizontal line: constant velocity; sloping up: accelerating; sloping down: decelerating.
Grade 5
Find acceleration from a velocity–time graphAcceleration = gradient = change in velocity ÷ time taken.
Quick recall
Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.
A bus accelerates from 4.0 m/s to 13 m/s in 6.0 s. Calculate the acceleration of the bus. Use the equation: acceleration = change in velocity ÷ time taken
1.5 m/s2
A trolley of mass 2.0 kg accelerates uniformly from rest to 3.0 m/s in 4.0 s. Calculate the acceleration of the trolley.
0.75 m/s2
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy4 marks
(a) A bus accelerates from 4.0 m/s to 13 m/s in 6.0 s. Calculate the acceleration of the bus. Use the equation: acceleration = change in velocity ÷ time taken[2]
(b) What is the acceleration of an object falling freely near the Earth’s surface?[1]
(c) What is meant by an object decelerating?[1]
Show the answer and mark scheme
(a)Answer: 1.5 m/s2
a = (13 − 4.0) ÷ 6.0
1.5 (m/s2)
(b)Answer: 9.8 m/s2
9.8 (m/s2)
(c)Answer: It is slowing down.
it is slowing down / its velocity is decreasing
Question 2Medium6 marks
(a) A cyclist accelerates uniformly from 2.0 m/s to 8.0 m/s over a distance of 40 m. Calculate the acceleration of the cyclist. Use the Physics Equations Sheet.[3]
(b) A stone is dropped from rest from a bridge 20 m above a river. Air resistance is negligible. acceleration due to gravity = 9.8 m/s2 Calculate the speed of the stone when it reaches the water. Use the Physics Equations Sheet.[3]