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.

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.

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

Section of graph | Gradient | Velocity | Acceleration |
---|---|---|---|

A | positive | increasing | positive |

B | zero | constant | zero |

C | negative | decreasing | negative |

D (v = 0) | zero | stationary (at rest) | zero |

- Question
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

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.

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: **

Ā½ Ć base Ć height

Ā½ Ć 4 Ć 8 = 16 m^{2}

**2. Find the area of the rectangle: **

(10 ā 4) Ć 8 = 48 m^{2}

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

(16 + 48) = 64 m