Answer
$\mu = \frac{v^2}{gr} - tan(tan^{-1}(\frac{v_0^2}{rg}))$
Work Step by Step
We find the angle of the road based on the maximum posted speed. Thus, we obtain:
$\theta = tan^{-1}(\frac{v_0^2}{rg})$
We also know:
$ mgcos\theta \mu+mgsin\theta=\frac{mv^2}{r}$
Thus, we find an expression for $\mu$:
$\mu=\frac{\frac{v^2}{r}-gsin\theta}{gcos\theta}$
$\mu = \frac{v^2}{gr} - tan\theta$
Plugging in the value of theta gives:
$\mu = \frac{v^2}{gr} - tan(tan^{-1}(\frac{v_0^2}{rg}))$
