I have two mathematical arguments (or, well, maybe the first one is mainly a physics argument) for why it is often unwise to speed.

Argument 1 (Quadratic braking distance)

The first argument is that the kinetic energy at speed v for a car with mass m is

    \[ KE = \frac{1}{2}mv^2. \]

The important part is that it scales quadratically with the velocity v. So if you drive twice as fast (i.e., replace v with 2v) you get four times the energy:

(1)   \[ KE_{\mathrm{double speed}} = \frac{1}{2}m(2v)^2 = \frac{1}{2}m2^2v^2 = \frac{1}{2}m4v^2 = 4\cdot\frac{1}{2}mv^2 = 4\cdot KE  \]

and thus your braking distance would also be quadrupled, giving you way less of a chance to react.

A concrete example, since 70 \approx 50 \cdot \sqrt{2}, if you drive 70kmph in a 50kmph zone, your energy and thus braking distance would be doubled (because \sqrt{2}^2 = 2), and if both you (driving 70kmph) and another car driving 50kmph encounter a tree on the road, and the other car manages to just barely brake in time, you will hit the tree with a velocity of 50kmph. (Assuming you have exactly the same car model, etc.) Now, replace the tree with a kid and it’s the difference between life and death.

Note that the first deceleration of 20kmph takes the same distance as the remaining 50kmph deceleration. This follows directly from (1), but intuitively this occurs because the faster you are driving, the longer distance you cover while decelerating.

Argument 2 (Harmonic Average)

The second argument concerns more so the utility of driving too fast, and is an interesting consequence of the AM-HM inequality.

Consider the scenario where you need to drive 100 km. For the first 50 km you are stuck in traffic so you are able to drive 50kmph. Then, for the second half, you can drive up to 150kmph (the speed limit might be 120) so in order to get to your destination faster, you drive 150kmph. Then you believe that you will average out so your average speed is

    \[ \frac{150 + 50}{2} = 100 \]

kmph. However, this is NOT the case, and in fact, it follows from the AM-HM inequality (which I will show in a bit) that your average speed is always lower than this simple “average.” But let’s first understand this example. The first 50 km section where you drive 50kmph takes 1 hour. And then the second 50 km distance takes 20 minutes since you drive thrice as fast. In total, the trip of 100 km takes 1 hour and 20 minutes (1.33… hours), giving you an average speed of

    \[ \frac{100\mathrm{km}}{1.33\mathrm{h}} = 75\mathrm{kmph}. \]

Yup. Just 75kmph. Not 100kmph like initially expected. Which means you took on “quadratically as much danger” (see Argument 1) but be scammed out the “benefit” (having a fast average speed, and thus arriving quicker) of it. Which just means that speeding is a case that is loaded against you in both ways. Surprisingly, even if you drive \inftykmph in the second half, your average speed would only be 100kmph! (See below for why.)

Now, if you’re curious about what I meant with AM-HM inequality, let me explain. In the case above, if you wanted to compute the average speed “correctly”, you actually need to use the so-called Harmonic Mean (as opposed to the Arithmetic Mean which is the normally used “average”). The harmonic mean of two (positive) numbers a and b is

    \[ HM = \frac{2}{\dfrac{1}{a} + \dfrac{1}{b}} \]

so the harmonic mean of the speeds 50kmph and 150kmph is

    \[ \frac{2}{\dfrac{1}{50} + \dfrac{1}{150}} = \frac{2}{\dfrac{3}{150} + \dfrac{1}{150}} = \frac{2}{\dfrac{4}{150}} = \frac{2\cdot150}{4} = \frac{300}{4} = 75 \]

as we also found out manually before. If you’re curious how to calculate the average speed if the two sections are not of equal length, you just set each numerator in the denominator to the appropriate proportion. E.g., if the two sections are 70% and 30% each, you’d calculate the average speed as

    \[ \frac{1}{\dfrac{0.7}{50} + \dfrac{0.3}{150}} = 62.5. \]

The fact of the matter is that it’s always the case that

    \[ AM \geqslant HM \]

where AM is the Arithmetic Mean, i.e.,

    \[ AM = \dfrac{a+b}{2}. \]

So, no matter what, you are always “scammed.”1 And we can see that even if you drive \inftykmph in the second half, your average speed is (we can’t really work with \infty in an equation but whatever):

    \[ \frac{2}{\dfrac{1}{50} + \dfrac{1}{\infty}} = \frac{2}{\dfrac{1}{50} + 0} = \frac{2}{\dfrac{1}{50}} = \frac{2\cdot50}{1} = 100 \]

kmph, as I also mentioned before. If this seems unintuitive to you, don’t worry, because it allegedly did to Einstein too.

</rant>

  1. You could counter this by saying that it is sunk cost fallacy to look at the previous traffic when deciding how fast to drive in the second half, and that we should always do what is optimal in our current situation, so our overall average speed doesn’t matter. But then, there are other arguments (such as Argument 1) for why this is unwise. ↩︎

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