**Speed** is the measurement of distance travelled over a unit of time. The distance travelled is divided by the time taken to travel that distance: 30 metres travelled in 2 seconds is `frac(30)(2)` = 15 metres per second.

Because the time is dividing into the distance, the units are also shown as dividing: the units are shown as `frac(m)(s)`, or ms^{-1}.

For transport and other uses, the speed is shown as mph or kph, where mph is miles per hour, and kph is kilometres per hour.

The formula for speed is `text(speed) = frac(text(distance))(text(time))`, or `S = frac(D)(T)`.

Speed can vary on a journey: speed will be slower in town than on a motorway. The **Average Speed** is the speed between two points over a given amount of time even although the vehicle may have been going faster or slower at different parts of the journey.

Note for Higher students: **Velocity** has both a speed AND a direction (it is a vector).

A lorry travels 215 miles for Guildford to Manchester. The journey takes 5 hours. What is the average speed of the lorry?

`text(speed)` | `= frac(text(distance))(text(time))` | |

Substitute: | `s` | `= frac(215)(5)` |

Divide both sides by 5: | `s` | `= 43` |

Answer: 43mph

The same lorry continues from Manchester to go onto Penrith. If the driver can maintain the same speed, and leaves at 2.40pm, when can he expect to arrive in Penrith, which is 107.5 miles away?

`text(speed)` | `= frac(text(distance))(text(time))` | |

Substitute: | `43` | `= frac(107.5)(t)` |

Multiply both sides by `t`: | `43t` | `= 107.5` |

Divide both sides by 43: | `t` | `= 2.5` |

Add 2.5 hours (2 hours 30 minutes) to 2.40pm to get 5.10pm

Answer: 5.10pm

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