 Distance-Time Graphs Worksheets | Questions and Revision | MME

# Distance-Time Graphs Worksheets, Questions and Revision

Level 4-5

## Distance-Time Graphs

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:

## Distance time graphs –  Key thinks to remember:

1) The gradient of the line = speed

2) A flat section means no speed (stopped)

3) The steeper the graph the greater the speed

4) Negative gradient = returning to start point (coming back) KS3 Level 4-5

## Check out the MME Learning Portal

Online exams, practice questions and revision videos for every GCSE level 9-1 topic!

No fees, no trial period, just totally free access to the UK's best GCSE maths revision platform.

KS3 Level 4-5

## Using Distance Time Graphs

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 $30$ km away from their starting point and that took them $2$ hours.

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

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

Part D: $\bf{12:30 - 14:00}$, they travelled the full $60$ km back to where they began. Calculate the speed – for each part of the journey:

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

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

$\text{Speed}= \dfrac{30}{2} = 15$ km/h

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

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

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

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

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

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

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

KS3 Level 4-5
KS3 Level 4-5

## Example: Bike Ride

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 $17:45$?

[3 marks] 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 $25$ km away from home, stopped briefly, and then travelled $25$ km back home. Therefore, she travelled $50$ km 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 $5$ km up to $25$ km, meaning she travelled $20$ km in total – this is the “change in $y$”. So, we get

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

KS3 Level 4-5

## GCSE Maths Revision Cards

(252 Reviews) £8.99

### Example Questions

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

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

On a graph, this looks like: 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:

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

Period 3:

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

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

(a) Distance travel is,

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

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

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,

$\text{Maximum speed } = 48$ km/h

### Worksheets and Exam Questions

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

Level 1-3 New Official MME