1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Tough tough kinematics

  1. Mar 13, 2012 #1
    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 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.
     
  2. jcsd
  3. Mar 13, 2012 #2
    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Tough tough kinematics
Loading...