Homework Help: Tough tough kinematics

1. Mar 13, 2012

xzibition8612

1. The problem statement, all variables and given/known data

A speeder zooms past a parked police car at a constant speed of 70 mph. Then, 3 sec later, the policewoman starts accelerating from rest at 10 ft/s^2 until her velocity is 85 mph. How long does it take her to overtake the speeding car if it neither slows down nor speeds up?

2. Relevant equations
x''=a
x'=at+c1
x=(a/2)t^2+c1t+c2

3. The attempt at a solution
Speeder velocity = 102.7 ft/s

Police equations of motion while accelerating:
x''=10
x'=10t
x=5t^2

Thus 124.7=10t ---> t=12.47 s

Since police starts moving 3 secs later, after 15.47s:
Speeder has covered 15.47 * 102.7 = 1588.8 ft
Police has covered 5(12.47)^2 = 777.5 ft

The distance between them is 811.3 ft.

From this point on both are constant velocity:

Police: x'=124.7 x=124.7t
Speeder: x'=102.7 x=102.7t+811.3

Equating the two equations and solving get t=36s.

Book says 52.6 sec. No idea where i went wrong.

2. Mar 13, 2012

tal444

You are forgetting the fact that the police car waited for 3 sec and accelerated for 12.47s, for a total of 15.47s. You also rounded your final answer quite a bit. I get 36.86s. Add 36.86 + 15.47 = 52.6s