Pearson Edexcel · A-Level · Mathematics
Course version: 9MA0 · Version from Mar 2026
Understand, use and interpret graphs in kinematics for motion in a…
Full specification wording
Kinematics: 7.2 Understand, use and interpret graphs in kinematics for motion in a straight line: displacement against time and interpretation of gradient; velocity against time and interpretation of gradient and area under the graph.
For straight-line motion, a displacement–time graph plots displacement on the vertical axis against time.
Checked against the Pearson Edexcel 9MA0 specification (opens in a new tab) on .
- Point 2 of 5
- 4 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
For straight-line motion, a displacement–time graph plots displacement on the vertical axis against time. Its gradient gives velocity: a positive gradient means motion in the positive direction, a negative gradient means motion in the negative direction, and a horizontal line means the particle is at rest. A velocity–time graph plots velocity vertically against time. Its gradient gives acceleration, since acceleration is the rate of change of velocity. The area between the graph and the time axis gives displacement, with areas above the axis counted positive and areas below counted negative; the total distance travelled is the sum of the magnitudes of these areas.
For example: A velocity rising steadily from 0 m s⁻¹ to 10 m s⁻¹ over 5 s has acceleration 2 m s⁻² and displacement equal to the triangular area ½ × 5 × 10 = 25 m.
What earns the marks
- Interpret the gradient of a displacement–time graph as velocity, including its sign as direction.
- Interpret the gradient of a velocity–time graph as acceleration.
- Interpret the area under a velocity–time graph as displacement, treating areas below the axis as negative.
- Distinguish displacement from distance travelled by summing signed areas for displacement and magnitudes for distance.
- Use graph features such as intercepts, turning points and horizontal sections to describe the motion in words.
Common mistakes and how to put them right
Reading the height of a velocity–time graph as displacement.
Put it right: Displacement is the area under the graph, while the height gives velocity.
Treating area below the time axis as positive when finding displacement.
Put it right: Such area contributes negatively to displacement but positively to distance travelled.
Confusing the gradient of a displacement–time graph with acceleration.
Put it right: This gradient gives velocity, while acceleration comes from the gradient of a velocity–time graph.
Exam tips
- Label axes with quantity and unit before sketching or reading a graph.
- Split the area under a velocity–time graph into simple shapes such as triangles, rectangles and trapezia.
- State the sign convention and use it consistently when interpreting gradients and areas.
Try the lesson
The start of the Kinematics: 7.2 Understand, use and interpret graphs in kinematics for motion in a straight line: displacement against time and interpretation of gradient; velocity against time and interpretation of gradient and area under the graph. lesson, made from checked teaching notes for this specification point.
Have a go first
On a displacement–time graph for motion in a straight line, a straight line rises steadily from the origin. What does the gradient of this line represent?
Listen
For straight-line motion, a displacement–time graph plots displacement on the vertical axis against time. Its gradient gives velocity: a positive gradient means motion in the positive direction, a negative gradient means motion in the negative direction, and a horizontal line means the particle is at rest. A velocity–time graph plots velocity vertically against time. Its gradient gives acceleration, since acceleration is the rate of change of velocity. The area between the graph and the time axis gives displacement, with areas above the axis counted positive and areas below counted negative; the total distance travelled is the sum of the magnitudes of these areas. For example, a velocity rising steadily from 0 m s⁻¹ to 10 m s⁻¹ over 5 s has acceleration 2 m s⁻² and displacement equal to the triangular area ½ × 5 × 10 = 25 m. A strong answer earns marks for these points. Interpret the gradient of a displacement–time graph as velocity, including its sign as direction. Interpret the gradient of a velocity–time graph as acceleration. Interpret the area under a velocity–time graph as displacement, treating areas below the axis as negative. Distinguish displacement from distance travelled by summing signed areas for displacement and magnitudes for distance. A common slip: Reading the height of a velocity–time graph as displacement. Instead, displacement is the area under the graph, while the height gives velocity. Exam tip: Label axes with quantity and unit before sketching or reading a graph.
See it
Key words
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Gradient as a rate of change
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Area under a rate graph
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Signed displacement versus total distance
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Graphical description of motion
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See every definition
- Gradient as a rate of change
- On a displacement–time graph the gradient gives velocity, and on a velocity–time graph the gradient gives acceleration, since acceleration is the rate of change of velocity.
- Area under a rate graph
- The area between a velocity–time graph and the time axis gives displacement, with areas above the axis counted positive and areas below counted negative.
- Signed displacement versus total distance
- Displacement is found by summing signed areas, while total distance travelled is the sum of the magnitudes of these areas.
- Graphical description of motion
- Graph features such as intercepts, turning points and horizontal sections are used to describe the motion in words, where a horizontal line means the particle is at rest.
Label the diagram
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1Box 1
2Box 2
3Box 3
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