Written by Colin+ in ask uncle colin.

Dear Uncle Colin,

I read that when cars are driving at 70mph on the motorway, they take up more space than when they travel more slowly (because you need to leave a longer safe gap between them). What’s the most efficient speed for motorway travel if you want to get as many cars past a given point as possible?

That’ll Really Annoy Fast Folk In Cars

Hi, TRAFFIC, and thanks for your message! What a lovely problem.

Let’s suppose – seeing as we’re mathematicians, not transport planners – that cars ought to keep the minimum stopping distance as recommended in the Highway Code between them.

That’s actually a nice quadratic sequences problem in itself, especially if you use miles per hour and feet1.

Speed (mph) | Stopping distance (ft) |
---|---|

20 | 40 |

30 | 75 |

40 | 120 |

50 | 175 |

60 | 240 |

70 | 315 |

If you fit a model2 of the form $D_f = Av_{mph}^2 + Bv_{mph} + C$ to that, you find $D_f = \frac{1}{20}v_{mph}^2 + v_{mph}$. Splendid. But we really don’t like those units: $D_f$ is in feet and $v_{mph}$ is in miles per hour and just ugh3.

We measure distance in metres, and there are about $\frac{10}{3}$ feet in one of those, so $D_f = \frac{10}{3}D$.

We measure speed in metres per second, and (it turns out) 9 miles per hour is about the same as 4 m/s. So, $v_{mph} = \frac{4}{9} v$

Rewriting the formula: $\frac{10}{3}D = \frac{1}{20} \times \frac{16}{81}v^2 + \frac{4}{9}v$

Or, simplifying, $D = \frac{2v^2 + 90v}{675}$. Much nicer.

In terms of how much space a car takes up, we also need to add on its length – and let’s say the average car is 4m long.

So, a car travelling at $v$ m/s takes up $D_+ = \frac{2v^2 + 90v}{675} + 4$ metres - and I’ll pretend the car and its associated stopping distance simply a box that long.

Suppose I’m standing by the motorway4 and timing how long the box of the car takes to pass me. The front of the box needs to travel $D_+$ metres, it’s travelling at $v$ m/s, so it will take $T=\frac{D_+}{v}$ seconds to pass me.

That gives $T = \frac{2v + 90}{675} + \frac{4}{v}$ seconds.

To maximise the number of cars going past a fixed point over any period of time, we need to minimise $T$ as a function of $v$. Let’s differentiate!

$\diff Tv = \frac{2}{675} - \frac{4}{v^2} = 0$ for an extremum5.

So, $v^2 = 1350$ and $v \approx 36.7$ m/s, which is 82.7 mph!

This is very naive analysis, of course, and just the first model that popped into my head – I don’t recommend driving at 82mph any more than I recommend standing by a motorway!

Hope that helps,

- Uncle Colin

* Many thanks to @realityminus3 for helpful comments and improving my work, as always.

- I’d normally not touch either with a three-metre pole [↩]
- I’m using subscripts to denote the imperial units, which we’ll change in a minute. [↩]
- One reason for the ugh is that if we try to differentiate this, it goes wrong: we've got two different distance units and they don't play nicely together. [↩]
- I would not do such a thing unless my car had broken down, obviously [↩]
- And, because of the shape of the curve, it's a minimum [↩]