AQA GCSE Physics

Revision Notes

5.6.12 Area Under Velocity-Time Graphs

Higher Tier Only

Area Under Velocity-Time Graphs

  • The area under a velocity-time graph represents the displacement (or distance travelled) by an object

Velocity-Time Area graph, downloadable IGCSE & GCSE Physics revision notes

The displacement, or distance travelled, is represented by the area beneath the graph

  • If the area beneath the graph forms a triangle (i.e. the object is accelerating or decelerating), then the area can be determined by using the following formula:

Area = ½ × Base × Height

  • If the area beneath the graph forms a rectangle (i.e. the object is moving at a constant velocity), then the area can be determined by using the following formula:

Area = Base × Height

Higher Tier Only

Determining Distance

  • Enclosed areas under velocity-time graphs represent total displacement (or total distance travelled)

Determining Distance on a V-T graph, downloadable IGCSE & GCSE Physics revision notes

Three enclosed areas (two triangles and one rectangle) under this velocity-time graph represents the total distance travelled

  • If an object moves with constant acceleration, its velocity-time graph will comprise of straight lines
    • In this case, calculate the distance travelled by working out the area of enclosed rectangles and triangles as in the image above
  • If an object moves with changing acceleration, its velocity-time graph will be a curve
    • In this case, estimate the distance travelled by counting the squares underneath the curve

Worked Example

The graph below shows how the velocity of a train changes with time as the train travels along a straight section of the journey.

WE Estimating Distance question image, downloadable IGCSE & GCSE Physics revision notes

Estimate the distance travelled by the train along this section of the journey.

Step 1: Identify whether distance can be determined exactly or by estimation

    • The train is moving with changing acceleration because the velocity-time graph is a curve
    • This means the distance should be estimated by counting squares underneath the curve

Step 2: Determine how much distance is represented by each square on the graph

WE Estimating Distance answer image, downloadable IGCSE & GCSE Physics revision notes

    • The area of each square on the graph can be calculated by using the equation:

Area = Base × Height

Area = 100 × 5 = 500

    • Therefore, the distance represented by each square is 500 m

Step 3: Write down the total number of squares underneath the curve

    • There are approximately 17 squares underneath the curve (including non-whole squares)

Step 4: Determine the total estimated distance 

    • The total estimated distance is found by multiplying the number of squares by the distance represented by each square:

Total estimated distance = Number of squares × Distance represented by each square

Total estimated distance = 17 × 500 = 8500 m

Worked Example

The velocity-time graph below shows a car journey which lasts for 160 seconds.

Area Under a V-T graph question, downloadable IGCSE & GCSE Physics revision notes

Calculate the total distance travelled by the car on this journey.

Step 1: Recall that the area under a velocity-time graph represents the distance travelled

    • In order to calculate the total distance travelled, the total area underneath the line must be determined

Step 2: Identify each enclosed area

    • In this example, there are five enclosed areas under the line
    • These can be labelled as areas 1, 2, 3, 4 and 5, as shown in the image below:

Area Under a V-T graph solution, downloadable IGCSE & GCSE Physics revision notes

Step 3: Calculate the area of each enclosed shape under the line

    • Area 1 = area of a triangle = ½ × base × height = ½ × 40 × 17.5 = 350 m
    • Area 2 = area of a rectangle = base × height = 30 × 17.5 = 525 m
    • Area 3 = area of a triangle = ½ × base × height = ½ × 20 × 7.5 = 75 m
    • Area 4 = area of a rectangle = base × height = 20 × 17.5 = 350 m
    • Area 5 = area of a triangle = ½ × base × height = ½ × 70 × 25 = 875 m

Step 4: Calculate the total distance travelled by finding the total area under the line

    • Add up each of the five areas enclosed:

total distance = 350 + 525 + 75 + 350 + 875

total distance = 2175 m

Exam Tip

Whenever you are asked to determine the distance using a velocity-time graph, start by stating that the distance = the area under the graph.

A common mistake is to try and find distance by using the equation for speed, distance and time – but this equation will not work if the object’s speed is changing. In this case, you can only estimate the distance travelled by approximating the area under the graph (counting squares).

Author: Jonathan

Jonathan graduated with a first-class Master's degree in Theoretical Physics from Imperial College London. He has worked in education for more than a decade as a Maths and Physics Teacher, Tutor, Head of Physics, and most recently, as Assistant Headteacher. He is now an Educational Consultant and works with us to design and improve our Physics resources.
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