Answer
Since the speed was $23.0~mi/h$ at the moment the brakes were first applied, the police officer can not give a ticket for speeding in a 25 mi/h zone.
Work Step by Step
We can find the rate of deceleration:
$F_N~\mu_k = ma$
$mg~\mu_k = ma$
$a = g~\mu_k$
$a = (9.80~m/s^2)(0.60)$
$a = 5.88~m/s^2$
We can find the initial velocity:
$v_f^2 = v_0^2+2ax$
$v_0^2 = v_f^2-2ax$
$v_0 = \sqrt{v_f^2-2ax}$
$v_0 = \sqrt{0-(2)(-5.88~m/s^2)(9.0~m)}$
$v_0 = 10.29~m/s$
We can convert the speed to units of mi/h:
$v_0 = 10.29~m/s\times \frac{1~mi}{1609~m} \times \frac{3600~s}{1~hr} = 23.0~mi/h$
Since the speed was $23.0~mi/h$ at the moment the brakes were first applied, the police officer can not give a ticket for speeding in a 25 mi/h zone.
