**Features of Time-Distance Graphs**

Graphs comparing distance and time should be called **Time-Distance graphs** because:

- Time is the independent variable so
**Time is marked on the horizontal x-axis**. - Distance is the dependent variable so
**Distance is marked on the vertical y-axis**.

**Calculating the Speed**

- To calculate the speed, select a
**triangular section of the graph**. In that section, find the distance travelled and the time this took. Then work out the speed (Distance divided by Time). - The steeper the slope of the graph, the faster is the speed.
- When the object is stationary (not moving), the graph is horizontal and has no slope.

A **hydrocopter** is a rescue vehicle has an aircraft engine and a catamaran hull (two hulls). It is amphibious (can travel over water or land/snow/ice).

The legs of a recent rescue journey are as follows:

- From its base, the hydrocopter travels 12 kilometres over ice-covered land in 12 minutes.
- Then, it travels 20 kilometres along a fiord (water) in 10 minutes to the capsized boat.
- The rescue of the distressed fisherman takes 5 minutes. During this time, the hydrocopter remains stationary.
- The rescue hydrocopter then retraces the same journey back to the base, but because it is carrying the rescued fisherman, its speed is slower. The journey over water takes 12 minutes and the journey over land takes 14 minutes.

LEG | TIME | DISTANCE |

1st (land) | 12 min | 12 km |

2nd (water) | 10 min | 20 km |

3rd (rescue) | 5 min | 0 km |

4th (water) | 12 min | 20 km |

5th (land) | 14 min | 12 km |

(a) Calculate the hydrocopter's speed (in kilometres per minute) at each of the five stages of the journey. Write the speed information in a table.

(b) Draw a distance-time graph showing all the legs of the journey.

**Answer (a):**

LEG | TIME | DISTANCE | SPEED |

1st (land) | 12 min | 12 km | 12 ÷ 12 = 1 km/min |

2nd (water) | 10 min | 20 km | 20 ÷ 10 = 2 km/min |

3rd (rescue) | 5 min | 0 km | 0 km/min (not moving) |

4th (water) | 12 min | 20 km | 20 ÷ 12 =1.7 km/min |

5th (land) | 14 min | 12 km | 12 ÷ 14 = 0.9 km/min |

**Answer (b):**

Because the time-distance line graph requires a **cumulative total of time on the x-axis** and **distance from the base on the y-axis**, a new table is needed.

CUMULATIVE TIME | DISTANCE FROM BASE |

0 min | 0 km |

12 min | 12 km |

22 min | 32 km |

27 min | 32 km |

39 min | 12 km |

53 min | 0 km |