# Ask Uncle Colin: Traffic Flow

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.

### Stopping distances

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.

### How long does it take to pass?

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.

### Now for some calculus!

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. ## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

1. I’d normally not touch either with a three-metre pole []
2. I’m using subscripts to denote the imperial units, which we’ll change in a minute. []
3. 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. []
4. I would not do such a thing unless my car had broken down, obviously []
5. And, because of the shape of the curve, it's a minimum []

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