## College Physics (4th Edition)

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.
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.