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Area under a velocity–time graph

Key idea: Find displacement and distance travelled from signed areas under velocity–time graphs, with an exam method and worked O-Level examples.

  • SEC G3 Physics 2027
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Learning objectives

  • State what speed means
  • State what velocity means, including its direction
  • Calculate average speed from total distance and total time
  • Calculate acceleration as change in velocity divided by time taken
  • State what uniform acceleration means
  • Interpret examples of non-uniform acceleration
  • Plot and interpret displacement–time and velocity–time graphs in one dimension
  • Deduce rest and uniform or non-uniform velocity from a displacement–time graph
  • Deduce rest, uniform velocity and uniform or non-uniform acceleration from a velocity–time graph
  • Use signed area under a velocity–time graph to determine displacement
  • Recall constant free-fall acceleration near Earth as approximately 10 m/s²

1. Definitions

The signed area under a velocity–time (v–t) graph gives the object’s displacement during that time interval.

Δ s = A_signed

Here, A_signed is the area between the graph and the time-axis, with its sign retained. Area above the axis is positive; area below it is negative. Choose the positive direction before calculating.

Signed area under a velocity–time graphA continuous velocity–time graph crosses the time axis. The region above the axis is labelled positive displacement and the region below is labelled negative displacement. Two summary boxes distinguish displacement from total distance travelled.Keep the sign of each regiontimevelocitypositive areaforward displacementnegative areamotion in the opposite directionv = 0: direction changesDisplacementpositive area − below-axis magnitudeThe result can be positive, negative or zero.Total distance travelledpositive area + below-axis magnitudeDistance is non-negative.
Scroll diagram horizontally to read all labels.
Area above the time axis is positive displacement; area below it is negative displacement. Add signed areas for displacement, but add their magnitudes for total distance travelled.

2. Key Ideas

  1. Mark the time interval required.
  2. Split the region into rectangles, triangles or trapeziums.
  3. Calculate each area with its sign.
  4. Add the areas and give the answer in metres.

Useful shapes are:

rectangle: A = vt; triangle: A = 1/2 × base × height; trapezium: A = (1/2)(v₁ + v₂)t

The units confirm the result: (m s⁻¹)(s) = m.

Velocity–time graph split into two shapes

Velocity is 5 metres per second from zero to 3 seconds, then decreases steadily to zero at 7 seconds. The displacement is a rectangle plus a triangle.

Scroll across the graph to read all labels.

Velocity is 5 metres per second from zero to 3 seconds, then decreases steadily to zero at 7 seconds. The displacement is a rectangle plus a triangle.Velocity is 5 metres per second from zero to 3 seconds, then decreases steadily to zero at 7 seconds. The displacement is a rectangle plus a triangle.
From 0–3 s use a rectangle; from 3–7 s use a triangle.
Open full-size graph
View figure data
Values for Velocity–time graph split into two shapes
Time (s)Velocity
05
35
70

3. Detailed Explanations

For constant velocity, Δ s = vΔ t. On a v–t graph, this product is the area of a rectangle. For a changing velocity, divide the interval into smaller strips and add their areas. Straight-line exam graphs let you combine the strips as familiar shapes.

The gradient has a different meaning: the gradient of a v–t graph gives acceleration.

4. Common Mistakes

  • Treating area below the time axis as positive distance when the question asks for signed displacement.
  • Using base × height for a triangular or trapezoidal region without its correct geometric factor.
  • Reading a value from the graph without stating the scale and units of both axes.

5. Exam Tips

  • Split the graph at every change of gradient and every crossing of the time axis.
  • Write the signed area of each simple shape with units before adding; take absolute values only when total distance is requested.
  • Check that an area from velocity in m/s and time in s produces metres.

6. Worked Examples

Modelled example 1

Rectangle and triangle

Core

Problem

The graph above shows an object travelling at 5.0 m s⁻¹ for 3.0 s, then slowing uniformly to rest over 4.0 s. Find its displacement.
Study the worked solution

  1. Split the region into familiar shapes

    Method

    Use a rectangle for the constant-velocity interval and a triangle for the slowing interval.

    Reason

    The signed area under a velocity–time graph gives displacement.

    Working

    A_rectangle = (5.0)(3.0) = 15 m

  2. Add the triangular area

    Reason

    The velocity falls linearly from 5.0 m s⁻¹ to zero over 4.0 s.

    Working

    Δ s = 15 + (1/2)(4.0)(5.0) = 15 + 10 = 25 m

Guided practice 2

Displacement from a trapezium

About 4 min

Problem

An object’s velocity increases uniformly from 2.0 m s⁻¹ to 8.0 m s⁻¹ over 4.0 s. Find its displacement.

Use the average height of the trapezium

Unit: m

Hints

Hint 1: identify the shape
The region between the line and time axis is a trapezium.
Hint 2: use parallel sides
Calculate (1/2)(2.0 + 8.0)(4.0).
View solution step by step

  1. Use the trapezium area

    Method

    Average the initial and final velocities, then multiply by time.

    Reason

    Uniformly changing velocity makes a straight-sided trapezium under the graph.

    Working

    Δ s = (1/2)(v₁ + v₂)t = (1/2)(2.0 + 8.0)(4.0) = 20 m

Common misconception 3

Crossing the time-axis

Find and correct the mistake

Learner response

Velocity falls uniformly from + 4.0 m s⁻¹ at 0 s to -2.0 m s⁻¹ at 6.0 s. A student adds both triangular areas as positive and says displacement equals distance. Locate the error and find both quantities.

Separate signed area from path length

Unit: m
Unit: m

View solution step by step

  1. Locate the axis crossing

    Method

    Split the graph where velocity becomes zero.

    Reason

    The line falls by 1.0 m s⁻¹ each second, reaching zero at 4.0 s.

    Working

    A₊ = (1/2)(4.0)(4.0) = 8.0 m, A₋ = -(1/2)(2.0)(2.0) = -2.0 m

  2. Calculate the two requested quantities

    Method

    Add signed areas for displacement and magnitudes for distance.

    Reason

    Displacement retains direction; distance records total path length.

    Working

    Δ s = 8.0-2.0 = 6.0 m, d = 8.0 + 2.0 = 10 m

Examiner practice 4

Piecewise velocity–time area

4 marks

Examination question

An object travels at 6.0 m s⁻¹ for 2.0 s, then slows uniformly to rest over the next 3.0 s. Find its total displacement. [4 marks]

Show the split, both areas and total

View solution step by step

  1. Identify displacement as area

    1 mark

    Method

    Use the region under the velocity–time graph.

    Reason

    Velocity–time area has unit metres and gives displacement.

    Working

    Split the region into a rectangle and a triangle.

  2. Calculate the rectangle

    1 mark

    Reason

    Velocity is constant for the first 2.0 s.

    Working

    A₁ = (6.0)(2.0) = 12 m

  3. Calculate the triangle

    1 mark

    Reason

    The velocity then falls linearly to zero over 3.0 s.

    Working

    A₂ = (1/2)(3.0)(6.0) = 9.0 m

  4. Add the signed areas

    1 mark

    Method

    Add the two above-axis regions.

    Reason

    Both intervals have positive velocity.

    Working

    Δ s = 12 + 9.0 = 21 m

Challenge 5

Work backwards from an area

Minimal support

Reverse area calculation

A velocity–time graph is a straight line from rest to velocity V at t = 6.0 s. The displacement is 30 m. Find V.

Use the known area to infer the graph height

Unit: m s^-1

Hints

Hint 1: identify the region
The area is a triangle with base 6.0 s and height V.
Hint 2: make height the subject
Solve 30 = (1/2)(6.0)V.
View solution step by step

  1. Equate area to displacement

    Method

    Use the given 30 m as the triangular area.

    Reason

    The graph begins at zero velocity and rises along a straight line.

    Working

    30 = (1/2)(6.0)V

  2. Solve for the final velocity

    Reason

    The unknown graph height is the velocity at 6.0 s.

    Working

    V = 10 m s⁻¹

Further mistakes to diagnose

  • Calculating the gradient instead of the area.
  • Treating area below the axis as positive when finding displacement.
  • Calling signed displacement “distance”. Distance travelled cannot be negative.
  • Omitting the time scale or velocity scale when finding a shape’s dimensions.
  • Giving the area unit as m s⁻¹ instead of m.

7. Mind Stretchers

Mind stretcher 1: Same displacement, different distanceExtension

A velocity–time graph has a positive area of 18 m and a negative area of -6 m. Find the displacement and distance travelled.

Show answer

For displacement, retain the signs:

Δ s = 18 + (-6) = 12 m

For distance travelled, add the magnitudes:

d = 18 + 6 = 24 m

Mind stretcher 2: Reconstruct a graph durationExtension

An object moves at constant velocity 4.0 m s⁻¹ for 3.0 s, then slows uniformly to rest over time T. Its total displacement is 20 m. Find T.

Show answer

The rectangular area is:

A₁ = (4.0)(3.0) = 12 m

The triangular area must therefore be 20-12 = 8 m:

8 = (1/2)(T)(4.0)

T = 4.0 s

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027