Stopping Distance for UK Driving Test

One of the exams that must be passed for a UK driving licence is driving theory. This is a series of multiple choice questions. Most of them are very obvious, but some questions require knowledge. In particular, the stopping distances for a car travelling at various speeds is required to be known. (In reality, one does not actually need to know the answers as guessing suffices, and so few (if any) questions are asked on it that it is permissible to get them wrong.)

The stopping distance is split into ‘thinking distance’ and ‘braking distance’. The official numbers are as follows:

 Speed (mph) Thinking distance (m) Stopping distance (m) Total distance (m) 20 6 6 12 30 9 14 23 40 12 24 36 50 15 38 53 60 18 55 73 70 21 75 96

One interesting thing about these numbers it that the speed is measured in miles per hour, but the stopping distance is in metres. One mile is 1.6 kilometres.

There is no need to memorise these numbers as it is simple to make an approximate formula for them.

The thinking distance should be the speed multiplied by a reaction speed, and this reaction speed should not depend on the speed the car is travelling at. That is, the thinking distance divides by the speed should be a constant. It is, and this constant is 0.3. Thus to calculate the thinking distance, multiply the speed by 0.3. (In practice, one divides the speed by 10 and then multiplies it by three)

For the stopping distance, the force applied by the brakes should not depend (much) on the speed that the car is travelling. Therefore if a(t) represents the acceleration at time t, it should satisfy $a(t) = -b$, where b is some unknown constant. Integrating, this means $v(t) = S-bt$, where v(t) represents the velocity at time t, and S is the initial speed. It thus follows that $s(t) = St - t^2$, where s(t) represents the position of the car at time t. The car stops at time S/b, and so the distance travelled at that point is $s(S/b) = S^2/b - S^2/b^2 = S^2( b - 1 )/b^2$.

We can solve that equation as a quadratic in b, given the speed S and the actual stopping distance s(S/b). Taking the larger root and trying it on the data points on the table above, we find b is approximately 64 (although it does vary between data points). Thus the stopping distance is $\frac{S^2}{64^2} (64-1) \approx \frac{S^2}{64}$. This is good because finding a 64th of something is simple, as you can just repeatedly divide by two.

This means that altogether, the total stopping distance in metres is:

$0.3S + \frac{S^2}{64}$

This gives the following results:

 Speed (mph) Actual total distance (m) Predicted total distance (m) 20 12 12.25 30 23 23.06 40 36 37.00 50 53 54.06 60 73 74.25 70 96 97.56

Which is close enough! (Indeed, it exactly gets these points if one divides by S/60 and takes the nearest whole number)

Note that although the equation is very possible to evaluate by hand, it could be easier. If one assumes that you will only be asked about speeds which are a multiple of ten, then you can let S = 10k. Then the equation becomes the following which is much easier to use:

$3k + 1.5 k^2$