Distance-Time Graphs Worksheets | Questions and Revision | MME

Distance-Time Graphs Worksheets, Questions and Revision

Level 4 Level 5

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:

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 $30$km 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 $30$km away from their home.

Part D: $\bf{12:30 - 14:00}$, they travelled the full $60$km 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 $25$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

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

Example Questions

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.

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.

(a) Distance travel is,

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

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

(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,

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

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