Gradients and area under graphs | Pearson Edexcel GCSE Maths Notes

Visual model

Area under a speed-time graph gives distance

distance = areatimespeed

Gold-standard guide

26 mins

What you will learn

Use graph shape to estimate rates and quantities.

Use a clear step-by-step method for gradients and area under graphs.

Check your answer and avoid the most common exam mistake.

Useful before you start

Core number skillsEarlier algebra skillsShowing clear working

Core knowledge

Know the rule, then use it

These are the short notes. Read each one, then check you can use it in the worked example below.

Method

Gradient of a velocity-time graph = acceleration

Step 1

Find acceleration from the gradient

Acceleration = change in velocity / change in time = (24 − 0) / (8 − 0) = 3 m/s²

Step 2

Identify the shape under the graph

From 0 to 8 seconds the graph is a straight line from (0, 0) to (8, 24), forming a triangle

Step 3

Calculate area of the triangle

Area = $\frac{1}{2}$ × base × height = $\frac{1}{2}$ × 8 × 24 = 96 m

Watch out

Students confuse which quantity the gradient and which quantity the area represent

Gradient

$gradient = rate of change.$

Area under speed-time

$area = distance travelled.$

Worked example

A velocity-time graph rises uniformly from 0 to 24 m/s in 8 seconds. Find the acceleration and the total distance travelled.

Find acceleration from the gradient: Acceleration = change in velocity / change in time = (24 − 0) / (8 − 0) = 3 m/s².

Identify the shape under the graph: From 0 to 8 seconds the graph is a straight line from (0, 0) to (8, 24), forming a triangle.

Calculate area of the triangle: Area = $\frac{1}{2}$ × base × height = $\frac{1}{2}$ × 8 × 24 = 96 m.

State units clearly: Distance = 96 m. Acceleration = 3 m/s².

Final answer

Acceleration = 3 m/s²; distance = 96 m

Question ladder

Build up to the hardest questions

Do them in order. If you miss a step, read the solution, then redo the question without looking.

Workedreasoning

A velocity-time graph rises uniformly from 0 to 24 m/s in 8 seconds. Find the acceleration and the total distance travelled.

4 marks4 minsgradients-and-area-under-graphs-worked

Show solution

Worked solution

1.Find acceleration from the gradient: Acceleration = change in velocity / change in time = (24 − 0) / (8 − 0) = 3 m/s².
2.Identify the shape under the graph: From 0 to 8 seconds the graph is a straight line from (0, 0) to (8, 24), forming a triangle.
3.Calculate area of the triangle: Area = $\frac{1}{2}$ × base × height = $\frac{1}{2}$ × 8 × 24 = 96 m.
4.State units clearly: Distance = 96 m. Acceleration = 3 m/s².

Final answer

Acceleration = 3 m/s²; distance = 96 m

Mark points

M1: find acceleration from the gradient
M1: identify the shape under the graph
M1: calculate area of the triangle
M1: state units clearly
A1: Acceleration = 3 m/s²; distance = 96 m

Watch out

Students confuse which quantity the gradient and which quantity the area represent.Remember: gradient = rate of change (here, acceleration). Area = accumulated quantity (here, distance).Never swap these two.

Diagnosticrecall

A v-t graph is horizontal at 15 m/s for 10 seconds. Find distance.

1 mark2 minsgradients-and-area-under-graphs-q1

Show solution

Worked solution

1.Spot the skill: Gradient of a velocity-time graph = acceleration.
2.Use the find acceleration from the gradient stage first, then identify the shape under the graph.
3.Keep the final answer visible: 150 m.

Final answer

150 m

Mark points

M1: use the correct gradient of a velocity-time graph = acceleration. area under a velocity-time graph = distance.for a straight-line section: area is a triangle or trapezium.estimate with counts of squares when the graph is curved.
A1: 150 m

Watch out

Easyprocedure

A v-t graph drops from 20 m/s to 0 in 5 s. Find deceleration.

2 marks3 minsgradients-and-area-under-graphs-q2

Show solution

Worked solution

1.Spot the skill: Gradient of a velocity-time graph = acceleration.
2.Use the identify the shape under the graph stage first, then calculate area of the triangle.
3.Keep the final answer visible: 4 m/s².

Final answer

4 m/s²

Mark points

M1: use the correct gradient of a velocity-time graph = acceleration. area under a velocity-time graph = distance.for a straight-line section: area is a triangle or trapezium.estimate with counts of squares when the graph is curved.
A1: 4 m/s²

Watch out

Mediumreasoning

A trapezium under a v-t graph has parallel sides 8 m/s and 14 m/s and width 6 s. Find distance.

3 marks4 minsgradients-and-area-under-graphs-q3

Show solution

Worked solution

1.Spot the skill: Gradient of a velocity-time graph = acceleration.
2.Use the calculate area of the triangle stage first, then state units clearly.
3.Keep the final answer visible: 66 m.

Final answer

66 m

Mark points

M1: use the correct gradient of a velocity-time graph = acceleration. area under a velocity-time graph = distance.for a straight-line section: area is a triangle or trapezium.estimate with counts of squares when the graph is curved.
A1: 66 m

Watch out

Hardproblem solving

On a distance-time graph, what does a decreasing gradient indicate?

3 marks5 minsgradients-and-area-under-graphs-q4

Show solution

Worked solution

1.Spot the skill: Gradient of a velocity-time graph = acceleration.
2.Use the state units clearly stage first, then find acceleration from the gradient.
3.Keep the final answer visible: The object is slowing down.

Final answer

The object is slowing down

Mark points

M1: use the correct gradient of a velocity-time graph = acceleration. area under a velocity-time graph = distance.for a straight-line section: area is a triangle or trapezium.estimate with counts of squares when the graph is curved.
A1: The object is slowing down

Watch out

Exam-stylemulti-step

A v-t graph rises from 4 m/s to 10 m/s in 3 s. Find acceleration and distance.

4 marks6 minsgradients-and-area-under-graphs-q5

Show solution

Worked solution

1.Spot the skill: Gradient of a velocity-time graph = acceleration.
2.Use the find acceleration from the gradient stage first, then identify the shape under the graph.
3.Keep the final answer visible: Acceleration = 2 m/s²; distance = 21 m.

Final answer

Acceleration = 2 m/s²; distance = 21 m

Mark points

M1: use the correct gradient of a velocity-time graph = acceleration. area under a velocity-time graph = distance.for a straight-line section: area is a triangle or trapezium.estimate with counts of squares when the graph is curved.
A1: Acceleration = 2 m/s²; distance = 21 m

Watch out

Grade 9 stretchproblem solving

A velocity-time graph rises uniformly from 4 m/s to 10 m/s over 6 seconds. Find the distance travelled.

4 marks7 minsgraphs-g9

Show solution

Worked solution

1.The area is a trapezium.
2.Use $\frac{1}{2}$ × (sum of parallel sides) x width.

Final answer

42 m

Mark points

M1: use trapezium area
M1: $\frac{1}{2}$ × (4 + 10) x 6
A1: 42 m

Watch out

Do not rush straight into arithmetic. Select the relevant method and show a complete chain of working.

Timed checkpoint

16 mins - 9 marks

Switch between skills

Set a timer and attempt all four questions before opening any answers. This is closer to the way skills appear in a real paper.

1Gradients and area under graphs - 2 marksA v-t graph is horizontal at 15 m/s for 10 seconds. Find distance.Mark answer

Answer

150 m

2Algebraic notation and substitution - 2 marksFind t² + 4t when t = −2Mark answer

Answer

−4

3Simplifying expressions - 2 marksSimplify 2xy + 3x²y − xy + x²yMark answer

Answer

3x²y + xy

4Expanding and factorising - 3 marksFactorise fully 6x² + 9xMark answer

Answer

3x(2x + 3)

Mastery check

I can explain the method for gradients and area under graphs.
I can show clear working without skipping key steps.
I can avoid this mistake: Students confuse which quantity the gradient and which quantity the area represent.Remember: gradient = rate of change (here, acceleration). Area = accumulated quantity (here, distance).Never swap these two.

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