Solution 2

We are still along the same stretch of road, with a speed limit of \(\quantity{50}{mph}\), and the average speed cameras set at \(\quantity{4}{miles}\) apart. In this instance our first driver spends the same amount of time driving at \(\quantity{46}{mph}\) as he does at \(\quantity{62}{mph}\). Our second driver drives the first half of the distance at \(\quantity{46}{mph}\) and the second half at \(\quantity{62}{mph}\).

What are the average speeds of each driver? Will either of them be caught speeding?

Distance time plot for the first driver.

Let’s start with the first driver, who travels half the time at \(\quantity{46}{mph}\) and half the time at \(\quantity{62}{mph}\). This is represented by the graph above.

We have two different ways to think about the journey; as a whole, and split into its two pieces.

We can write the speeds of the two pieces as \[46=\dfrac{p}{\frac{1}{2}t}\quad\text{and}\quad 62 = \dfrac{q}{\frac{1}{2}t},\] where \(p\) and \(q\) are the distances travelled.

The journey as a whole can be written as \[ \text{average speed} = \dfrac{p+q}{t}.\]

We know that \(p+q = 4\) but it turns out that substituting this into the equation is not helpful.

Therefore we can substitute in values for \(p\) and \(q\) in terms of \(t\), \[ \textrm{ average speed } = \dfrac{23t+31t}{t},\] which leaves us with an average speed of \(\quantity{54}{mph}\).

What do you notice about the average speed? Could you have reasoned through this answer more easily, without doing the calculations?

Now we come to the driver who travels half the distance at each speed. Again we can think about the parts of the journey and the journey as a whole.

Distance time plot for the second driver.

The separate parts can be written \[46 = \dfrac{2}{t_1} \quad \textrm{and} \quad 62 = \dfrac{2}{t_2}.\]

Will they spend a longer time travelling at \(\quantity{46}{mph}\) or at \(\quantity{62}{mph}\)?

The whole journey gives us

\[\text{average speed}=\dfrac{4}{t_1+t_2}.\]

Rearranging the above equations to find \(t_1\) and \(t_2\) and then substituting these in gives

\[\text{average speed}=\dfrac{4}{\dfrac{2}{46}+\dfrac{2}{62}}= \quantity{52.8}{mph}\ \text{($3$ sf)}.\]

Do you think that either driver will be given a speeding ticket?

What happens if the speed cameras are \(\quantity{10}{miles}\) apart instead? What changes?

Did you find that anything changed as the distance increased? Does this surprise you? Can you prove your findings for any distance?

If we look back at the equations we solved for the first driver, when the times were equal, we had \(p+q\) as the distance, which we knew equalled \(4\), but we did not need to use this fact.

Whatever the distance, we would have the same first set of equations \[46=\dfrac{p}{\frac{1}{2}t}\quad \textrm{and}\quad 62 = \dfrac{q}{\frac{1}{2}t},\] and the same second equation \[\text{average speed} = \dfrac{p+q}{t},\] where \(p+q=\) total distance travelled.

We could rearrange these to write \[\dfrac{\dfrac{46t}{2}+\dfrac{62t}{2}}{t} = \dfrac{46+62}{2}=54,\] so the distance does not affect the average speed.

It is the same in the second case, although perhaps less obvious. If instead of \(4\) miles we have \(2d\) miles, then we get the equations \[46 = \dfrac{d}{t_1} \quad \text{and}\quad 62 = \dfrac{d}{t_2}\] and \[\text{average speed}=\dfrac{2d}{t_1+t_2}.\]

Rearranging the first two equations in terms of \(t_1\) and \(t_2\) means we we can write this as \[\text{average speed}=\dfrac{2d}{\dfrac{d}{46}+\dfrac{d}{62}}\] and simplify it to \[\text{average speed}= \dfrac{2}{\dfrac{1}{46}+\dfrac{1}{62}}= \quantity{52.8}{mph}\text{ (3 sf)}.\]

Therefore this average also does not depend on the distance travelled.

Can you also explain graphically why the average speed for each situation does not depend on the total distance?

We have calculated two different average speeds. Often when people say ‘average’ they mean one particular type. Think about other averages you have met - do either of these two calculations remind you of averages you already know?

For equal distance spent at each speed

\[\textrm{average speed} = \dfrac{2}{\dfrac{1}{46}+\dfrac{1}{62}},\]

and for equal time spent at each speed

\[\textrm{average speed} = \dfrac{46+62}{2}.\]

To think more about this go to the Averages section.