Motion - AQA SynergyVelocity-time graphs

The movement of objects can be described using motion graphs and numerical values. These are both used to help in the design of faster and more efficient vehicles.

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Velocity-time graphs

Watch this video to see how a velocity-time graph can be used to find out if an object is accelerating, at a constant velocity, or decelerating.

James Stewart explains how to find acceleration from a velocity-time graph

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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.

, \(\text{v}\) = 28 m/s

, \(\text{u}\) = 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.

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
Section of graphA
GradientPositive
VelocityIncreasing
AccelerationPositive
Section of graphB
GradientZero
VelocityConstant
AccelerationZero
Section of graphC
GradientNegative
VelocityDecreasing
AccelerationNegative
Section of graphD (v = 0)
GradientZero
VelocityStationary (at rest)
AccelerationZero

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)

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
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