Those maniacs!

The likelihood of a car passing you (or you passing it) is directly proportional to its speed. You will never be passed by a car going your speed. Intuitively, if there were a bunch of cars going 1 and 2 km/h faster than you, twice as many 2's would pass you.

So we can multiply the speed distribution by $kx$ to get a function that models the proportion of cars passing you by speed. For simplicy, use a standard normal distrbution:

$p(x) = kx \cdot \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^{2}}$

A plot of this function shows what we'd expect. For $x>0$ there is a bulge of cars going a fair amount faster than you. Even though most cars are goiong nearly your speed, you don't see them much. Even though very fast cars will pass you, there are fewer of them. The maximum appears to be at 1 standard deviation. Let's confirm this:

$p'(x)=\frac{k}{\sqrt{2\pi}} \left(e^{-\frac{1}{2}x^{2}}-x^{2}e^{-\frac{1}{2}x^{2}}\right)$

Setting $p'(x)=0$

$(1-x^{2})e^{-\frac{1}{2}x^{2}}=0$

$x=\pm 1$

So the car passing you most are going one standard deviation faster than you, or $\boxed{107}$ km/h.

Also, the cars you pass most are only going $93$