GCSEScienceOCRCombined Science A (Gateway)PhysicsRevision NotesForcesMotionArea Under Velocity-Time Graphs

Area Under Velocity-Time Graphs (OCR GCSE Combined Science A (Gateway): Physics): Revision Note

Exam code: J250

Written by: Ashika

Reviewed by: Caroline Carroll

Updated on 14 September 2026

Area Under Velocity-Time Graphs

Higher Tier Only

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

Velocity–time graph showing distance equals the area beneath the graph: an initial triangle labelled “one half times base times height” followed by a rectangle labelled “base times height”; axes are velocity and time. — **The displacement, or distance travelled, is represented by the area beneath the graph**

If the area beneath the graph forms a triangle (e.g. the object is accelerating uniformly from rest or decelerating uniformly to rest), 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

Determining distance from a velocity-time graph

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

Velocity–time graph showing three enclosed areas beneath a line: Area 1 is an accelerating triangle, Area 2 a constant-velocity rectangle, and Area 3 a decelerating triangle. — **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 straight lines
- In this case, calculate the distance travelled by working out the area of enclosed rectangles and triangles as in the image above

Worked Example

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

Velocity–time graph of a car journey: velocity rises from 0 to 17.5 metres per second by 40 seconds, stays constant to 70 seconds, rises to 25 by 90 seconds, then falls to 0 at 160 seconds.

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

Answer:

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:

Annotated velocity–time graph identifying five areas: triangle Area 1 from 0–40 seconds, rectangle Area 2 from 40–70 seconds, split Areas 3 and 4 from 70–90 seconds, and triangle Area 5 from 90–160 seconds; velocity reaches 25 metres per second.

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

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