## What you need to know

Distance-time graphs are a way of **visually expressing a journey.** With distance on the y-axis and time on the x-axis, a distance-time graph tells us how far someone/something has travelled and how long it took them/it to do so.

Make sure you are happy with the following topics before continuing:

– Gradients of Straight Line Graphs

**Distance time graphs – Key thinks to remember: **

**1) The gradient of the line = speed**

**2) The steeper the graph the greater the speed**

**3) Flat section means no speed (stopped)**

**4) Negative gradient = returning to start point (coming back)**

The graph below describes a journey that has several parts to it, each represented by a different straight line.

**Part A**: \bf{09:00 - 11:00}, the person travelled 30km away from their home and that took them 2 hours.

**Part B**: \bf{11:00 - 12:00}, we can see that the line is flat, so their distance from home did not change – they were stationary.

**Part C**: \bf{12:00 - 12:30}, they moved a further 30km away from their home.

**Part D**: \bf{12:30 - 14:00}, they travelled the full 60km back to where they began.

**Calculating Speed**

Calculate the speed for each part of the journey:

\text{Speed} (S) = \dfrac{\text{Distance}(D)}{\text{Time}(T)}

**Part A**: \bf{09:00 - 11:00}

\text{Speed}= \dfrac{30}{1} = 30 \text{ km/h}

**Part B**: \bf{11:00 - 12:00}

\text{Speed}= \dfrac{0}{1} = 0\text{ km/h} (Not moving)

**Part C**: \bf{12:00 - 12:30}

\text{Speed}= \dfrac{30}{0.5} = 60 \text{ km/h}

**Part D**: \bf{12:30 - 14:00}

\text{Speed}= \dfrac{60}{1.5} = 40\text{ km/h}

From this we can see that the person travelled the fastest over **part C**

**Example**

Valentina is going for a bike ride. Below is a distance-time graph that describes her full journey.

a) How long was she stationary for?

b) What was the total distance travelled during her journey?

c) What was her average speed in kilometres per hour between 17:15 and 17:45?

a) We can see that the graph was flat for the duration of one big square. From the axis, we can see

that two big squares total 15 minutes, therefore one big square is worth 7.5 minutes, so she was stationary for 7.5 minutes.

b) Valentina travelled away 25km away from home, stopped briefly, and then travelled 25km back home. Therefore, she travelled 50km in total.

c) We need to calculate the gradient of the graph between 17:15 and 17:45. This period lasted for 30 minutes, which is equivalent to 0.5 hours – this is the “change in x”. During this period, she increased her distance from home from 5km up to 25km, meaning she travelled 20km in total – this is the “change in y”. So, we get

\text{Gradient } = \dfrac{20}{0.5} = 40\text{km/h}.

### Example Questions

**Question 1:** From the description of Neil’s journey below, construct a distance-time graph.

Neil left home at 12:00 and after half an hour of moving at a constant speed, he had travelled 12km. At 12:30 his speed increased, and after an hour of travelling at a constant speed he had travelled a further 32km, at which point he stopped. After 3 hours of being stopped, he drove towards home at a constant speed, and it took him 2 hours in total to get home.

So, making the words above seem more readable, we get:

- 12:00 – 12:30, he travels from 0km away to 12km away;
- 12:30 – 13:30, he travels from 12km away to 44km away;
- 13:30 – 16:30, he stays in one place;
- 16:30 – 18:30, he travels from 44km away to 0km away.

On a graph, this looks like:

Note that it might look like one straight line from 12:00 to 13:30, but it does actually change slightly at 12:30, so be careful with your drawings.

**Question 2**: Below is a distance-time graph describing a 1500m race run by Chris. Work out the maximum speed he reached during this run to 3 significant figures.

Looking at the graph, we can see that he runs at three different speeds during different portions of the race. The graph becomes less steep in the middle, so that won’t be his period of maximum speed, and the other two are hard to distinguish just by looking so we’ll work them both out.

Period 1:

\text{Gradient } = \dfrac{\text{distance travelled}}{\text{time taken}} = \dfrac{600 - 0}{72 - 0} = 8.33\text{m/s}

Period 3:

\text{Gradient } = \dfrac{\text{distance travelled}}{\text{time taken}} = \dfrac{1,500 - 880}{282 - 180} = 6.08\text{m/s}

Therefore, the fastest speed travelled by Chris during the race was 8.33m/s, to 3sf.

**Question 3: **A cyclist on a training ride records the distance she travels away from home. The data only shows the first 150 minutes of the ride before her cycling computer ran out of battery.

(a) How far did she ride in the first 150 minutes.

(b) For how long was she stationary,

(a) Distance travel is,

\text{Total Distance } = \48 km + 10 km = 58 km

(b) She stopped for 30 mins at the 32 km mark.

**Question 4: **A car journey on a day trip to the coast is recorded.

(a) Work out the maximum speed over the course of the trip.

(a) The gradient of a distance time graph is the speed. Hence to find the fastest average speed we must find the steepest section of the graph.

This is the final section which cover 48 km in one hour, thus,

\text{Maximum speed } = \8 \text{km/h}

**Question 5: **

### Worksheets and Exam Questions

#### (NEW) Distance Time Graphs Exam Style Questions - MME

Level 1-3#### Time Tables And Distance Tables - Drill Questions

Level 1-3#### Distance Time Graphs - Drill Questions

Level 4-5### Videos

#### Distance Time Graphs Q1

GCSE MATHS#### Distance Time Graphs Q2

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