GCSE Physics resources

AQA GCSE Forces

Motion Graphs, Speed and Acceleration

Use graph gradient and area carefully. A curved distance–time graph shows changing speed; a horizontal velocity–time line shows constant velocity; a negative velocity can still have a positive speed.

AQA 8463AQA 8464 GCSE PhysicsCombined Science: Trilogy FoundationHigher

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The revision snapshot

Use graph gradient and area carefully. A curved distance–time graph shows changing speed; a horizontal velocity–time line shows constant velocity; a negative velocity can still have a positive speed.

By the end, you should be able to:

  • Read the story in a graph.
  • Apply the forces model, evidence or method to an unfamiliar exam context.
  • Recognise and correct this wrong turn: Using area under a distance-time graph.

Before you revise

Diagnostic question

Choose an answer from memory. Your result tells you what to watch for in the guide.

Which statement correctly summarises motion graphs, speed, velocity and acceleration?

Core revision guide

Learn the model, then use it

Motion Graphs, Speed and Acceleration questions become manageable when the central model, evidence and exam method are kept together. Read the explanation, test the misconception and then apply the idea without looking back.

Read the story in a graph

Use graph gradient and area carefully. A curved distance–time graph shows changing speed; a horizontal velocity–time line shows constant velocity; a negative velocity can still have a positive speed.

Build the physics picture

Use the graph labels before choosing a rule. Distance-time gradient is speed; velocity-time gradient is acceleration; area under a velocity-time graph is displacement. A horizontal velocity-time line means constant velocity, while a line crossing zero shows a change of direction.

Relationships used in this guide

Speed: speed = distance ÷ time (m/s, m, s). Acceleration: a = Δv ÷ t (m/s², m/s, s). Choose each relationship from the physical change described, convert quantities into compatible units and keep the unit beside the final answer.

Distance-time and velocity-time graphs show gradient triangles and the area under a velocity-time graph.
Gradient gives speed or acceleration; area under a velocity-time graph gives displacement.Open the full-size exam diagram
Speedspeed = distance ÷ timem/s, m, s
Accelerationa = Δv ÷ tm/s², m/s, s

Misconception clinic

Replace the tempting answer

Read each belief, say what is wrong with it, then compare your correction.

Tempting idea: Using area under a distance-time graph.

Use instead: Use graph gradient and area carefully. A curved distance–time graph shows changing speed; a horizontal velocity–time line shows constant velocity; a negative velocity can still have a positive speed.

Tempting idea: Calling a horizontal velocity-time graph stationary.

Use instead: Use the graph labels before choosing a rule. Distance-time gradient is speed; velocity-time gradient is acceleration; area under a velocity-time graph is displacement. A horizontal velocity-time line means constant velocity, while a line crossing zero shows a change of direction.

Tempting idea: Ignoring negative velocity direction.

Use instead: Name the axis quantities, describe the gradient or area and translate it into a sentence about motion.

Retrieval practice

Quick checks

1. Which exam method is most reliable for motion graphs, speed, velocity and acceleration?

2. Which statement correctly applies motion graphs, speed, velocity and acceleration to a question?

3. A pupil writes: “Using area under a distance-time graph.” Which replacement is accurate?

Explore the relationship

Motion graph reader

Treat the gradient of this distance–time graph as speed, then calculate distance = gradient × time.

What to notice: use the readout to describe how one variable changes when the other is controlled.

Worked examples

See the method being built

These examples expose the difference between a secure read the story in a graph method and the common error “Using area under a distance-time graph.”.

Example 1 · Distance-time gradient

A runner travels 150 m in 30 s at constant speed. Find graph gradient.

  1. Use speed = distance ÷ time.
  2. Calculate 150 ÷ 30.
  3. Give m/s.
Show the answer

Gradient = 5.0 m/s.

Example 2 · Velocity-time acceleration

Velocity changes from 4.0 m/s to 16 m/s in 6.0 s. Find acceleration.

  1. Use change in velocity ÷ time.
  2. Calculate (16 − 4) ÷ 6.
  3. Give m/s².
Show the answer

Acceleration = 2.0 m/s².

Example 3 · Velocity-time area

A vehicle travels at 8.0 m/s for 5.0 s. Find displacement.

  1. Area is rectangle under graph.
  2. Multiply velocity by time.
  3. Give metres.
Show the answer

Displacement = 40 m.

Exam precision

Exam technique: Read the story in a graph

Do this: Name the axis quantities, describe the gradient or area and translate it into a sentence about motion.

Independent practice

Exam-style questions

Write an answer before opening the marking guidance. Then edit the exact phrase or step that would gain the next mark.

1. State what the gradient of a distance-time graph represents.

[1 mark]
Show marking guidance and model answer

Marking guidance: Award one mark for the precise statement shown in the model answer.

Model answer: Speed.

2. State what the gradient of a velocity-time graph represents.

[1 mark]
Show marking guidance and model answer

Marking guidance: Award one mark for the precise statement shown in the model answer.

Model answer: Acceleration.

3. State what area under a velocity-time graph represents.

[1 mark]
Show marking guidance and model answer

Marking guidance: Award one mark for the precise statement shown in the model answer.

Model answer: Displacement.

4. A distance-time graph is horizontal. Describe motion.

[2 marks]
Show marking guidance and model answer

Marking guidance: 2 marks are available for relevant, linked physics points that match the model answer.

Model answer: The object is stationary; distance is not changing.

5. A velocity-time line crosses zero. What does this show?

[2 marks]
Show marking guidance and model answer

Marking guidance: 2 marks are available for relevant, linked physics points that match the model answer.

Model answer: The object changes direction.

6. A car increases velocity from 0 to 20 m/s in 10 s. Calculate acceleration.

[3 marks]
Show marking guidance and model answer

Marking guidance: Award one mark for the correct relationship, one for a valid substitution and one for the final answer with its unit.

Model answer: a = 20 ÷ 10 = 2.0 m/s².

Questions pupils ask

Motion Graphs, Speed and Acceleration FAQs

What is the main idea in Motion Graphs, Speed and Acceleration?

Use graph gradient and area carefully. A curved distance–time graph shows changing speed; a horizontal velocity–time line shows constant velocity; a negative velocity can still have a positive speed.

What mistake should I avoid in Motion Graphs, Speed and Acceleration?

Using area under a distance-time graph. Use the graph labels before choosing a rule. Distance-time gradient is speed; velocity-time gradient is acceleration; area under a velocity-time graph is displacement. A horizontal velocity-time line means constant velocity, while a line crossing zero shows a change of direction.

Useful next steps

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