### Calculus meets Functions

Problem requiring decisions

## Solution

The Official Highway Code describes typical stopping distances for cars (section 126 of the 2015 edition). They are given as a table showing distances for different speeds of travel. Each stopping distance is made up of two parts – a thinking distance and a braking distance. They are summarised below.

Speed, $v$ Thinking distance Braking distance Stopping distance, $d_s$
$\quantity{20}{mph}$ $\quantity{6}{m}$ $\quantity{6}{m}$ $\quantity{12}{m}$
$\quantity{30}{mph}$ $\quantity{9}{m}$ $\quantity{14}{m}$ $\quantity{23}{m}$
$\quantity{40}{mph}$ $\quantity{12}{m}$ $\quantity{24}{m}$ $\quantity{36}{m}$
$\quantity{50}{mph}$ $\quantity{15}{m}$ $\quantity{38}{m}$ $\quantity{53}{m}$
$\quantity{60}{mph}$ $\quantity{18}{m}$ $\quantity{55}{m}$ $\quantity{73}{m}$
$\quantity{70}{mph}$ $\quantity{21}{m}$ $\quantity{75}{m}$ $\quantity{96}{m}$

One thing to notice and bear in mind is that we have some peculiarly mixed up units in the table. Speeds are quoted in $\mathrm{mph}$ or miles per hour, whereas distances are quoted in metres. Where it became necessary, we chose to use a conversion rate of $\quantity{1}{mile}\approx\quantity{1600}{m}$.

Look at the data in the table. What relationships do you see between the distances and how do they vary with speed?

Write an equation expressing the stopping distance, $d_s$, in terms of speed, $v$.

Firstly, note that the stopping distance is equal to the sum of the thinking and braking distances, as suggested in the introductory text.

The thinking distance goes up by $\quantity{3}{m}$ every time the speed goes up by $\quantity{10}{mph}$ so it’s reasonable to suggest this is a linear relationship such as $d_{think}=\frac{3}{10}v$ where the speed, $v$, is in $\mathrm{mph}$ and the distance is in $\mathrm{m}$.

A linear relationship implies that the time taken to think is a constant, independent of speed, which seems reasonable.

What is the actual thinking time used in this formula?

When the speed doubles, the braking distance increases by a factor of $4$, although the numbers are not exact. We presume this is due to rounding as the data was put into the table.

So the braking distance appears to be proportional to the square of the speed.

This is based on a model of the work done by the brakes.

• The work done in slowing the vehicle is “$\mathrm{force}\times\mathrm{distance}$”.
• If the brakes apply a constant force, the work done is proportional to the distance moved.
• The work required to stop the car is equal to its initial kinetic energy which depends on $v^2$.
• So the braking distance is proportional to the square of the speed.

Can we write an equation to express the relationship?

We want to match the data with an equation of the form $d_{brake}=k v^2$. We could substitute in pairs of values from the table and work out the value of $k$ for each one – we find they vary a bit because the numbers have been rounded. Or we could work out $v^2$ for each row in the table and find $k$ as the gradient of a line of best fit. Graphing software such as Desmos will enable us to get a pretty good estimate.

We found quite a good fit for $d_{brake}=\frac{3}{200}v^2$ which would mean the stopping distance, $$$d_s = \frac{3}{10}v + \frac{3}{200}v^2. \label{eq:stopping}$$$

Plot a graph of $d_s$ against $v$, for speeds between zero and $\quantity{70}{mph}$.

You could do this either from the equation or straight from the table of data.

This graph was drawn using the equation and it fits the data in the table reasonably well.

How many average car lengths is $\quantity{30}{m}$ or $\quantity{96}{m}$?

The Highway Code goes on to say that when driving you should leave a gap of at least the stopping distance between you and the vehicle in front.

It also says that in faster-moving traffic, you should leave a “two-second gap”. In other words the front of your car should not reach a fixed point on the road until at least two seconds after the rear of the previous vehicle passed the same point.

Write down an equation for this two-second distance, $d_t$, in terms of $v$ and add it to your graph of the stopping distance.

The length of the two-second gap will be directly proportional to the speed (if I drive twice as fast, I cover twice the distance in the same time). After sorting out the mixed up units, we found $$$d_t = \frac{8}{9}v. \label{eq:twosec}$$$

Up to this point the units could be ignored, but because we are given the time in seconds we can’t avoid doing some conversion here. We did it by writing the speed as $\quantity{v\times1600\div3600}{m\,s^{-1}}$.

A graph of $d_s$ and $d_t$ looks like this.

At $\quantity{60}{mph}$ which of the two distances is bigger? Why might the Highway Code make the two-second suggestion?

At what speeds is $d_t=d_s$?

Perhaps surprisingly, at higher speeds like $\quantity{60}{mph}$ the suggested separation of vehicles is the smaller of the two distances. This may be in recognition of the fact that in faster traffic the car in front is unlikely to come to a complete stop without itself taking time to slow down.

The two distances are the same when $v=0$ and when $v\approx\quantity{39}{mph}$. We can read this value off the graph, or we could set the two expressions $\eqref{eq:stopping}$ and $\eqref{eq:twosec}$ equal and solve the resulting quadratic.

Perhaps $\quantity{40}{mph}$ is the speed above which the Highway Code considers traffic to be “faster-moving”.

In a model of traffic flow on a single lane road, it is assumed that each vehicle is $\quantity{4}{m}$ long, travelling at constant speed and separated from the one in front by the typical stopping distance for that speed.

Find an expression for the rate of traffic flow, $R_s$, in vehicles per hour, as a function of the speed, $v$. Plot a graph of this function for speeds up to $\quantity{70}{mph}$. Use your graph or the algebra to find the minimum or maximum value of this function and at what speed(s) it occurs?

In this model, the distance between the front of one car and the front of the next is $\quantity{d_s+4}{m}$. The flow rate will be the speed (in metres per hour) divided by this distance (in metres), which works out to be $R_s = \frac{v\times1600}{\frac{3}{10}v+\frac{3}{200}v^2+4} = 1600\times200\frac{v}{3v^2+60v+800}.$

A graph of this function looks like this.

We can get a good estimate of the maximum flow rate from the graph.

Alternatively, it is possible to calculate it exactly by differentiating using the quotient rule. We found there is a stationary point at $v=20\sqrt{\frac{2}{3}}$.

The maximum flow rate is roughly $2025$ cars per hour if they all drive at about $\quantity{16.3}{mph}$.

Can you explain why the flow rate decreases as speeds increase beyond this?

If instead of the typical stopping distance, the vehicles are separated by the two-second rule, what is the flow rate, $R_t$? What would the maximum or minimum flow rate be and when does it occur?

This time, the flow rate $R_t = 1600\times9\frac{v}{8v+36}$ whose graph is a translated and stretched hyperbola.

The maximum flow rate is $1800$ cars per hour but to achieve it they’d have to be driving infinitely fast!

How much does the flow rate increase as the speed increases from $\quantity{60}{mph}$ to $\quantity{70}{mph}$?

Under either model, what happens when the speed is very small? How must you drive?

What happens when the speed is zero?

If we allowed the speed to be negative how could we stick to the two-second rule?

Why might these models be unrealistic in practice?

There are lots of unrealistic assumptions in the models we have used here.

• Every vehicle is assumed to be driving at an exactly constant speed.
• In practice it is hard to accurately judge distances.
• Even if you can, some drivers aren’t good at sticking to safe distances.
• The models take no account of vehicles joining or leaving the road.

How many more can you think of?