Velocity-time graphs

Determining acceleration

If an object moves along a straight line, its motion can be represented by a velocity-time graph. 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 graph represents:

Section of graphGradientVelocityAcceleration
APositiveIncreasingPositive
BZeroConstantZero
CNegativeDecreasingNegative
D (v = 0)ZeroStationary (at rest)Zero

Calculating displacement - higher

curriculum-key-fact
The displacement of 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)

Example

Calculate the total displacement of the object, whose 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 displacement can be found by calculating the total area of the shaded sections below the line.

Find the area of the triangle:

\frac{1}{2} \times base \times height

\frac{1}{2} \times 4 \times 8 = 16~m^{2}

Find the area of the rectangle:

base × height

(10 - 4) × 8 = 48 m2

Add the areas together to find the total displacement:

(16 + 48) = 64 m