AQA · GCSE · Physics
Course version: 8463 · Version from Sept 2016
(HT only) interpret enclosed areas in velocity–time graphs to determine…
Full specification wording
Acceleration: (HT only) interpret enclosed areas in velocity–time graphs to determine distance travelled (or displacement).
On a velocity–time graph, the area enclosed between the plotted line and the time axis represents the distance travelled or displacement.
Checked against the AQA 8463 specification (opens in a new tab) on .
- Point 19 of 23
- 3 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
On a velocity–time graph, the area enclosed between the plotted line and the time axis represents the distance travelled or displacement. To determine this area, split the space under the graph into simple geometric shapes: rectangles where velocity is constant, and triangles where velocity changes uniformly. For a rectangle, use area = base × height. For a triangle, use area = ½ × base × height. Add these separate areas together to obtain the total distance.
For example: If a car travels at 15 m/s for 4 s, then accelerates uniformly to 25 m/s over 6 s, the distance is the initial rectangle (15 × 4 = 60 m) plus the acceleration phase's rectangle (15 × 6 = 90 m) and triangle (½ × 6 × 10 = 30 m), giving a total distance of 180 m.
What earns the marks
- State that the area between the line and the time axis represents distance travelled or displacement.
- Divide the area under the graph into simple geometric shapes, such as rectangles and triangles, to facilitate calculation.
- Calculate the full area under an acceleration phase by including both the rectangular base and the triangular upper component.
- Add the separate calculated areas to find the total distance travelled, providing the final answer with the unit m.
Common mistakes and how to put them right
Omitting the rectangular area under an acceleration phase.
Put it right by ensuring the entire area down to the time axis is calculated, not just the triangle above the initial velocity.
Forgetting the ½ in the triangle area formula.
Put it right: Explicitly writing out area = ½ × base × height before substituting any values.
Using the gradient instead of the area to find distance.
Put it right by remembering that gradient represents acceleration, while the enclosed area represents distance or displacement.
Exam tips
- Sketch the graph and shade or label each simple shape down to the time axis before doing any arithmetic to avoid missing sections.
- Check the units on the axes before calculating; if time is given in minutes, convert it to seconds to match a velocity in m/s.
Try the lesson
The start of the Acceleration: (HT only) interpret enclosed areas in velocity–time graphs to determine distance travelled (or displacement). lesson, made from checked teaching notes for this specification point.
Have a go first
A velocity–time graph shows a straight horizontal line at 8 m/s from 0 s to 5 s. What distance does the object travel in this time?
Listen
On a velocity–time graph, the area enclosed between the plotted line and the time axis represents the distance travelled or displacement. To determine this area, split the space under the graph into simple geometric shapes: rectangles where velocity is constant, and triangles where velocity changes uniformly. For a rectangle, use area = base × height. For a triangle, use area = ½ × base × height. Add these separate areas together to obtain the total distance. For example, if a car travels at 15 m/s for 4 s, then accelerates uniformly to 25 m/s over 6 s, the distance is the initial rectangle (15 × 4 = 60 m) plus the acceleration phase's rectangle (15 × 6 = 90 m) and triangle (½ × 6 × 10 = 30 m), giving a total distance of 180 m. A strong answer earns marks for these points. State that the area between the line and the time axis represents distance travelled or displacement. Divide the area under the graph into simple geometric shapes, such as rectangles and triangles, to facilitate calculation. Calculate the full area under an acceleration phase by including both the rectangular base and the triangular upper component. Add the separate calculated areas to find the total distance travelled, providing the final answer with the unit m. A common slip: Omitting the rectangular area under an acceleration phase. Instead, ensuring the entire area down to the time axis is calculated, not just the triangle above the initial velocity. Exam tip: Sketch the graph and shade or label each simple shape down to the time axis before doing any arithmetic to avoid missing sections.
Key words
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Area under a graph as a physical quantity
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Compound geometric shapes in physics
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Distance and displacement from motion graphs
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See every definition
- Area under a graph as a physical quantity
- The area enclosed between the plotted line and the time axis on a velocity–time graph represents the distance travelled or displacement.
- Compound geometric shapes in physics
- The space under a graph is split into simple geometric shapes, rectangles where velocity is constant and triangles where velocity changes uniformly.
- Distance and displacement from motion graphs
- The separate calculated areas are added together to obtain the total distance travelled, giving the final answer with the unit m.
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