G5 | Maths Made Easy


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

a) how long she was stationary for

b) the total distance travelled during her journey

c) 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.