AQA GCSE Combined Science (8464), Higher tier · Physics › Forces › Forces and motion › Describing motion along a line
Practise Acceleration. 18 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, find acceleration from the gradient of a velocity–time graph and, on the Higher tier, find the distance travelled from the area under it. You also need free fall and terminal velocity. This subtopic produces many calculation questions.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
4
Calculate acceleration using a = Δv ÷ tFrom 4 m/s to 16 m/s in 3 s: a = 12 ÷ 3 = 4 m/s2.
4
Describe motion from a velocity–time graphHorizontal line: constant velocity; sloping up: accelerating; sloping down: decelerating.
5
Find acceleration from a velocity–time graphAcceleration = gradient = change in velocity ÷ time taken.
6
Use the equation v² − u² = 2asIt is on the equations sheet and applies to uniform acceleration; rearrange it for a, s, u or v.
6
Explain terminal velocity for a falling objectIt accelerates at first; drag increases with speed until the resultant force is zero.
7
Estimate the size of everyday accelerationsFor example, a car reaching 30 m/s in about 10 s accelerates at about 3 m/s2.
7
Find distance from the area under a v–t graphSplit the area into rectangles and triangles and add them.
8
Count squares to find area under curvesCount the squares under a curved line and multiply by the distance one square represents.
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
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.
Area under the graph = distance travelled (or displacement). Split the area into rectangles (base × height) and triangles (½ × base × height). grade 7+
For a curved line, count the squares under it (count a square if at least half of it is under the line), then multiply by the distance one square represents. grade 8+
Falling and terminal velocity
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.
Cheatsheet
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
v–t graph: area under the line = distance travelled grade 7+
Terminal velocity: drag = weight, resultant force = 0
How to answer each type of question
Calculate acceleration
2 to 3 marks4
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 answer
Δv = 20.0 − 8.0 = 12.0 m/s (1) a = 12.0 ÷ 40 (1) a = 0.30 m/s2 (1)
Use v² − u² = 2as
3 marks6
Write down u, v, a and s, including any that are zero.
Substitute into v2 − u2 = 2 a s. Use a negative acceleration for slowing down.
Rearrange and calculate. Remember the square root if you are finding a velocity.
Example. A car travelling at 25 m/s decelerates uniformly at 5.0 m/s2 until it stops. Calculate the distance it travels while decelerating.
Show the model answer
02 − 252 = 2 × (−5.0) × s (1) −625 = −10 × s (1) s = 62.5 m (1)
Gradient and area of a velocity–time graph
4 marks7
Acceleration: gradient = change in velocity ÷ time.
Distance: split the area under the line into rectangles and triangles.
Add the areas, with the unit m.
Example. A velocity–time graph for a cyclist is a straight line from 0 m/s at 0 s to 8.0 m/s at 10 s, then a horizontal line at 8.0 m/s until 30 s. (a) Calculate the acceleration in the first 10 s. (b) Calculate the total distance travelled in the 30 s.
Show the model answer
(a) a = 8.0 ÷ 10 (1) = 0.80 m/s2 (1) (b) triangle = ½ × 10 × 8.0 = 40 m; rectangle = 20 × 8.0 = 160 m (1) total distance = 200 m (1)
Shortcuts and memory tricks
Velocity–time graph: slope = acceleration, area = distance.
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.
Where marks are lost
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.
Leaving out the ½ for a triangle area, or using ½ for a rectangle.
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.
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'.
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 scooter accelerates uniformly at 2.0 m/s2 from rest for 5.0 s. Calculate its final velocity. Use the equation: acceleration = change in velocity ÷ time taken
10 m/s
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]
Show the answer and mark scheme
(a)Answer: 0.75 m/s2
8.02 − 2.02 = 2 × a × 40
60 = 80 a
a = 0.75 (m/s2)
(b)Answer: 19.8 m/s
v2 − 0 = 2 × 9.8 × 20
v2 = 392
v = 19.8 (m/s)
Question 3Hard9 marks
An aircraft must reach an airspeed (its speed relative to the air) of 72 m/s to take off. The aircraft accelerates uniformly from rest along the runway at 2.4 m/s2.
(a) On a day with no wind, calculate the minimum length of runway needed. Use the Physics Equations Sheet.[3]
(b) Calculate the time taken for the aircraft to reach take-off speed.[2]
(c) On another day a steady wind of 12 m/s blows towards the front of the aircraft (a headwind). The airspeed of the aircraft is then 12 m/s greater than its speed relative to the ground. The acceleration relative to the ground is still 2.4 m/s2. Calculate the minimum length of runway needed on this day.[3]
(d) Suggest why aircraft take off into the wind.[1]
Show the answer and mark scheme
(a)Answer: 1080 m
722 − 0 = 2 × 2.4 × s
5184 = 4.8 s
s = 1080 (m)
(b)Answer: 30 s
t = 72 ÷ 2.4
30 (s)
(c)Answer: 750 m
speed relative to the ground needed = 72 − 12 = 60 (m/s)
602 = 2 × 2.4 × s
s = 750 (m)
(d)Answer: They reach take-off airspeed in a shorter distance.
a shorter runway / less time is needed to reach the take-off airspeed