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Homework Help: Help Sos

  1. Nov 4, 2007 #1
    Help!!! Sos

    Sylvia was driving at the constant speed of 25 m/s. She missed the posted speed limit sign in a school zone she was passing through. A police car was hiding on a curve and police car started accelerating from rest at 2.0 m/s2. From the time that Sylvia’s car passes police car

    How long will it take the police car to catch us to Sylvia’s car?

    I just need to know how to solve this, fist try I got 12.5 seconds, but I am not sure if iam right, what i did was i divide 25 by 2. I really need help with this, can anyone please give me tips or hints please..... I was struggling with this for an hour T.T
  2. jcsd
  3. Nov 4, 2007 #2
    Can some one help me please...
  4. Nov 4, 2007 #3
    can someone help me please
  5. Nov 4, 2007 #4
    You need to break the problem up into two pieces: before the police car accelerates, and the time it takes to catch the speeder. In both pieces of the problem, each car has a constant acceleration, so that allows us to use the kinematic equations. I'm hoping that wasn't the exact wording of the problem. If you still have trouble, could you state the problem word by word, and tell me which concept you're having trouble with exactly? This will help us avoid wrong answers.
  6. Nov 4, 2007 #5
    ok i will try now and see if it works...
  7. Nov 4, 2007 #6
    I really dont know how to get this because from 5 variables, they only give you 2 for police car and 3 for Sylvia's car... Can you give me more tips?.. please and thankyou
  8. Nov 4, 2007 #7
    First, I'd like to know if the problem you posted was the exact wording of the problem, once I know that, it will be much easier to help you. From what you wrote, it sounds like the police car accelerates instantaneously, and this is typically not the case.
  9. Nov 4, 2007 #8
    yes it is exact word from the probllem, police car is accelerating from rest which is 0m/s instantaneously its accelerating..
  10. Nov 4, 2007 #9
    Does the problem mention something like "it takes 3 seconds for the police car to react" or "after 3 seconds the police car begins accelerating"? If it doesn't say something like that, it assumes the police officer steps on the gas the very moment the speeder passes him. Check one last time for me, then we'll tackle the problem :)
  11. Nov 4, 2007 #10
    It tells us that it steps on the gas from the time Sylvia's car passes police car..
  12. Nov 4, 2007 #11


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    Write the equation for Sylvia's car... displacement in terms of time... Sylvia's going at a constant velocity.

    write the equation for the police car... it has uniform acceleration, a = 2.
  13. Nov 4, 2007 #12
  14. Nov 4, 2007 #13
    Ok then, lets break it down. The speeder has zero acceleration, the police car has a constant acceleration (which is given). We have to assume, since the police car apparently accelerates right at the moment the speeder passes him, that no initial distance was covered by the speeder (which again is very strange). We know that when they catch eachother, their displacements will be equal. We use the displacement equation for constant acceleration: [tex] x = x_0 + v_0t + .5at^2 [/tex]. You'll have one for the police car, and one for the speeder, simply set them equal to eachother and solve for the time.
  15. Nov 4, 2007 #14
    D= 25+25/2 *t

  16. Nov 4, 2007 #15
    oh so i have to substitute the equation right?
  17. Nov 4, 2007 #16
    is acceleration of sylvia's car is 0?
  18. Nov 4, 2007 #17
    Well, you would set the two equations equal to each other. The distance equation for the police car looks fine, but how do you get 25 + 25/2*t for the speeder?

    Edit: Yes, sylvia moves at a constant velocity, therefore she has zero acceleration.
  19. Nov 4, 2007 #18
    oh my bad,, i used wrong formula hold on please...
  20. Nov 4, 2007 #19
    d=25*t+(t squared/2) is this right?...
  21. Nov 4, 2007 #20
    I assume you get (t squared/2) from the acceleration part of the equation, but think of this: if sylvia has 0 acceleration, what happens to 1/2 a t^2?
  22. Nov 4, 2007 #21
    oh yea... it becomes 0 .... hehe..
  23. Nov 4, 2007 #22
    So it will be just d=25t for Sylvia's
  24. Nov 4, 2007 #23
    Assuming she's covered zero distance previous to the police car accelerating, yes.
    Last edited: Nov 4, 2007
  25. Nov 4, 2007 #24
    d=0t+(2tsquared/2) and d=25t



    50t = 2t squared

    25t = t squared and

    t is 5?
  26. Nov 4, 2007 #25


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    careful with the math...

    [tex]25t = t^2[/tex]

    continue from here...
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