#### Answer

The proof is below.

#### Work Step by Step

We first find the velocity that would cause the cars to slide this far:
$\frac{1}{2}mv^2 = F_f \times \Delta x$
$\frac{1}{2}mv^2 = mg \mu \Delta x$
$v=\sqrt{2g \mu \Delta x}=13.1\ m/s=47.19 \ km/h$
This means that the initial momentum must have been:
$=(2140+1040)(47.19)=150,062$
The slowest possible speed scenario would be if the large car possessed most of this momentum. Thus, we find:
$150062 =\sqrt{(2140^2v^2+1040^2(55^2)}$
$v=64.82 \ km/h$
Thus, it is clear that at least one of the cars was speeding.