Velocity-time graphs

Acceleration

Acceleration is the rate of change of velocity. It is the amount that velocity changes per unit time.

The change in velocity can be calculated using the equation:

change in velocity = final velocity – initial velocity

\Delta \text{v} = \text{v - u}

The average acceleration of an object can be calculated using the equation:

\text{acceleration} = \frac{\text{change in velocity}}{\text{time taken}}

\text{a} = \frac{\Delta \text{v}}{\text{t}}

This is when:

  • acceleration ( \text{a}) is measured in metres per second squared (m/s2 )
  • change in velocity ( \Delta \text{v}) is measured in metres per second (m/s)
  • time taken ( \text{t}) is measured in seconds (s)

If an object is slowing down, it is decelerating (and its acceleration has a negative value).

Example

A car takes 8.0 s to accelerate from rest to 28 m/s. Calculate the average acceleration of the car.

final velocity, \text{v} = 28 m/s

initial velocity, \text{u} = 0 m/s (because it was at rest – not moving)

change in velocity, \Delta \text{v} = (28 – 0) = 28 m/s

\text{a} = \frac{\Delta \text{v}}{\text{t}}

= 28 ÷ 8

= 3.5 m/s2

Question

A car takes 25 s to accelerate from 20 m/s to 30 m/s. Calculate the acceleration of the car.

final velocity, \text{v} = 30 m/s

initial velocity, \text{u} = 20 m/s

change in velocity, \Delta \text{v} = (30 – 20) = 10 m/s

\text{a} = \frac{\Delta \text{v}}{\text{t}}

= 10 ÷ 25

= 0.4 m/s2

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

Scientists draw graphs of data to help analyse a situation. A velocity-time graph of a journey can give information about acceleration (the gradient) and distance travelled (displacement).

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)
curriculum-key-fact
The displacement of an object can be calculated from the area under a velocity-time graph.

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.

  1. Find the area of the triangle:
    • ½ × base × height
    • ½ × 4 × 8 = 16 m2
  2. Find the area of the rectangle:
    • base × height
    • (10 – 4) × 8 = 48 m2
  3. Add the areas together to find the total displacement:
    • (16 + 48) = 64 m