Velocity-time graphs

Determining acceleration

For a moving object, the velocity can be represented by a velocity-time graph.

A horizontal line on a velocity-time graph, shows that the object is at constant velocity, but a sloping line on a velocity-time graph, shows that the object is accelerating.

The gradient of the line is equal to the acceleration of the object.

A velocity/time graph. Graph with four distinct sections. All lines are straight.

The table shows what each section of the red line on the graph represents:

Section of graphGradientVelocityAcceleration
D (v = 0) zerostationary (at rest)zero
Y axis: velocity in m/s. X axis: time in s. Red line, object moves at constant acceleration. Green line object moves at constant acceleration, levels at constant velocity, then constant deceleration.

Calculate the acceleration of the object shown by the red line on the graph above.

Find the gradient of the red line:

change in velocity = (10 - 0) = 10 m

change in time = (2 - 0) = 2 s

α = \frac{v - u}{t}

α = 10 ÷ 2

α = 5 m/s^{2}

Calculating the distance travelled

The distance travelled by an object can be calculated from the area under a velocity-time graph.

The area under the graph can be calculated by:

  • using geometry (if the lines are straight)
  • counting the squares beneath the line (particularly if the lines are curved).


Calculate the total distance travelled by the object - its motion is represented by the velocity-time graph below.

The y axis shows velocity in metres per second and the x axis time in seconds.  The object increases its velocity from 0 metres per second to 8 metres per second in 4 seconds.

Here, the distance travelled can be found by calculating the total area of the shaded sections below the line.

1. Find the area of the triangle:

½ × base × height

½ × 4 × 8 = 16 m2

2. Find the area of the rectangle:

(10 – 4) × 8 = 48 m2

3. Add the areas together to find the total displacement:

(16 + 48) = 64 m