Physics please help asap?

Question

A speeding motorist traveling 110 km/h passes a stationary police officer. The officer immediately begins pursuit at a constant acceleration of 10.2 km/h/s (note the mixed units).
(a) How much time will it take for the police officer to reach the speeder, assuming that the speeder maintains a constant speed?
_________s
(b) How fast will the police officer be traveling at this time?
_________km/h

Answer 1

I disagree with hunted1820

a. If Vo=0 as in this case, the formula
Vf = Vo + a*t
converts to
Vf = a*t and
t = Vf/a
That's what hunted1820 tried to do. The numbers he used gets the officer up to 110 km/h. But that doesn't quite do it. The cop started out at 0 km/hr and it took over 10 sec to get up to 110. Meanwhile, the speeder is getting a ways out in front. The cop has to go faster than 110 to catch up.

Note that the 2 cars are side by side when the problem starts and side by side when it ends. So the cop car's average velocity is same as the speeder's: 110 km/hr. When acceleration is uniform the whole time, average velocity is calculated
Vave = (Vo + Vf)/2
where our cop's Vave = 110 km/hr and Vo = 0. If you work that equation, you will find that Vf = 2*Vave.

Since the acceleration is uniform, the cop car will accelerate to 2X the speeder's speed. (That's the way the problem is written anyway. The cop continues with the same acceleration the entire time. Actually the cop would probably level his speed off at about 150 km/hr and take a while longer to catch the speeder.) But as it's written, the cop will get up to 220 km/hr. So time would be
t = Vf/a
where Vf = 220 km/h.

b. See above.

Answer 2

(a) = (110 km/h)/(10.2 km/h/s)
-Plug it in your calculator, and the answer is in seconds, b/c the km/h cancels.

(b) = 110 km/h.....common sense