Answer
$v = 0.60c$
Therefore, Jill should be cited for exceeding the 0.5c speed limit.
Work Step by Step
Let $L_0 = 100~m$ and let $L = 80~m$.
We can find the speed:
$L = L_0~\sqrt{1-\frac{v^2}{c^2}}$
$\sqrt{1-\frac{v^2}{c^2}} = \frac{L}{L_0}$
$1-\frac{v^2}{c^2} = (\frac{L}{L_0})^2$
$\frac{v^2}{c^2} = 1-(\frac{L}{L_0})^2$
$v^2 = [1-(\frac{L}{L_0})^2]~\times c^2$
$v = \sqrt{1-(\frac{L}{L_0})^2}~\times c$
$v = \sqrt{1-(\frac{80~m}{100~m})^2}~\times c$
$v = \sqrt{1-(0.80)^2}~\times c$
$v = 0.60~c$
Therefore, Jill should be cited for exceeding the 0.5c speed limit.
