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Challenging police car question

  1. Sep 19, 2012 #1
    1. The problem statement, all variables and given/known data
    A policeman sitting on a side of the road sees a speeder drive by him at 40 m/s. At the same time the speeder passing the policeman 2 things happen:
    1. the policeman steps on the gas and begins to accelerate fowards at a rate of 2.5 m/s^2 until he reaches 35m/s
    2. the speeder hits the breaks and begins to slow down at a rate of 4 m/s^2 untill he reaches the speed limit of 25m/s

    A) How long will it take for the cop to catch the speeder?
    B) How far will the policeman have to travel before he reaches the speeder


    2. Relevant equations
    Vf=Vi+at
    x=Vi(t)+(.5)at^2
    Vf^2=Vi^2+2ax


    3. The attempt at a solution
    I tried to solve it by setting the distance that the speeder goes equal to that of the cop and just plugging in initial velocity, acceleration, and time for both, but I did not get the right answer
     
  2. jcsd
  3. Sep 19, 2012 #2
    I think the problem with that approach is that the speeder and the cop don't accelerate for equal amounts of time. Speeder decelerates until he gets to 25 m/s from 40 m/s, while the cop accelerates until he gets to 35 m/s from 0 m/s.
     
  4. Sep 19, 2012 #3

    SammyS

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    Hello jayjay12077. Welcome to PF !

    Exactly what did you plug into what?

    What were your results?

    How long does it take the patrolman to reach 35 m/s? How far does he travel in this time?

    How long does it take the speeder to slow to 25 m/s? How far does he travel in this time?

    What should you do after answering these questions?
     
  5. Sep 19, 2012 #4
    SammyS. Well i got that the cop would take the cop 14 sec. to reach his final velocity. In that time he travels 245 meters. The speeder takes 3.75 seconds to slow down to his final velocty and during that time he travels 121.875 meters. So i thought that since they each had to travel the same distance that I use the equation x=Vit+.5at^2 for the cop and then for the speeder and set the two equations equal to each other. But when I did that and simplifed I got 35t=25t and that does not make sense
     
  6. Sep 19, 2012 #5
    The values for time and displacements for the accelerated portions of the motion seem to be OK. The last part does not seem OK. How did you get this 25t=35t?
     
  7. Sep 20, 2012 #6

    SammyS

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    After 14 sec., the cop has a constant velocity (zero acceleration).

    After 3.75 sec., the speeder has a constant velocity (zero acceleration).

    Where is the speeder at 14 sec.?
     
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