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xzibition8612
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Homework Statement
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?
Homework Equations
x''=a
x'=at+c1
x=(a/2)t^2+c1t+c2
The Attempt at a Solution
Speeder velocity = 102.7 ft/s
Police velocity = 124/7 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.