**Distance-Time Graphs ***Revision and Worksheets*

## 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. Let’s see what one might look like.

So, this graph describes a journey that has several parts to it, each represented by a different straight line. During the first part, 09:00 – 11:00, the person travelled 30km away from their home and that took them 2 hours. During the next part, 11:00 – 12:00, we can see that the line is flat, so their distance from home did not change – they were stationary. Then, for the third part of the journey, 12:00 – 12:30, they moved a further 30km away from their home. In the final part, 12:30 – 14:00, they travelled the full 60km back to where they began.

In the third part of the journey, it took them half an hour to travel 30km, compared to the first part of the journey wherein it took them 2 hours to travel the same distance. Clearly, they must have been travelling at a much faster speed in third part than in the first part, which leads us into the other thing we can do with distance-time graphs: calculate the speed.

Firstly, recall that speed is equal to distance divided by time (more info on that here (https://mathsmadeeasy.co.uk/gcse-maths-revision/speed-distance-time-gcse-revision-and-worksheets/)). Secondly, recall that to calculate the gradient of a line we divide the change in y, which here is distance, by the change in x, which here is time (more info on that here (https://mathsmadeeasy.co.uk/gcse-maths-revision/gradients-straight-line-graphs-gcse-maths-revision-worksheets/)). Combining these two facts, we conclude that the gradient of a distance-time graph is equal to the speed travelled during that period. Indeed, we can see that the in the graph above the line is much steeper from 12:00-12:30 than it is from 09:00-11:00.

**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}.

**Note: **if you are asked to calculate the average speed over a longer period of time which contains several lines of different graphs, you still want to do the same calculation: divide the change in distance by the change in time. Even though you aren’t strictly calculating the gradient of one particular line, it’s still as if you’re looking at calculating an average gradient during that period.

In addition to this, you may be asked to construct your own distance-time graph from a description.

For example, if a question begins:

“Neil left home at 12:00 and after half an hour of moving at a constant speed, he travelled 12km…”

Then you should start your graph at 0 and 12:00, and then draw a line that ends at 12:30 on the time axis and 12km on the distance axis. See: right.

Perhaps you should have a go at the question below as it definitely maybe will begin a little like this example…

## Example Questions

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.

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 3sf.

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.

## Distance-Time Graphs Revision and Worksheets

## Distance-Time Graphs Teaching Resources

Distance time graph questions are often answered quite poorly even by talented students. It is essential that students have a lot of practice and make sure they get these marks in the bag as usually these are the more straight forward marks on the GCSE Maths papers. Regardless of if you teach Maths in Leeds or tutor GCSE Maths in York, you should find our Distance-Time teaching resources useful.