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Constant acceleration of skateboarder

  1. Sep 15, 2008 #1
    1. The problem statement, all variables and given/known data
    A skateboarder coasts down a long hill for 6.00 seconds with constant acceleration, and starting from rest, covers a distance x1. On a second trip down the same hlil starting fron rest, he coasts for 2.00 seconds, covering a distance x2. Find the ratio of x1 to x2.


    2. Relevant equations



    3. The attempt at a solution
    To me the obvious answer would be
    6s to 2s, so 3 to 1.
    But something tells me it's not that simple.
    What am I missing?
     
  2. jcsd
  3. Sep 16, 2008 #2

    Redbelly98

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    Welcome to Physics Forums southernbelle.

    Correct, it's not that simple. That 3:1 ratio would work for a constant velocity (no acceleration).

    What equations do you know about that deal with constant acceleration?
     
  4. Sep 16, 2008 #3
    I know that xf=xi +vi(t) + 1/2at^2
    and
    vf=vi + at

    So....
    x2=0 + 0t +1/2a(4) and x1=0 + 0t + 1/2a(36)
    would the acceleration be -9.8 because of gravity? but I guess not because then I would get a negative answer.
     
  5. Sep 16, 2008 #4

    Redbelly98

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    Looking good, you're almost there.

    You're right that acceleration is not -9.8 m/s^2, but that's okay. Try taking the ratio x1/x2, like the problem asks, and use those two expressions you got.
     
  6. Sep 17, 2008 #5
    Okay, so it would be:

    18a/2a.

    So that would be 9a/a.

    So vf=9a=a.
    I am stuck :(
     
  7. Sep 17, 2008 #6

    Redbelly98

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    Yes, so far so good. Can you simplify that fraction,

    [tex]
    \frac{9a}{a}
    [/tex]

    Huh? No, that doesn't follow from anything, plus they are not asking for vf.

    They are asking for the ratio of x1/x2, which you (correctly) got is 9a/a. You just need to simplify 9a/a, and then you're done.
     
  8. Sep 17, 2008 #7
    Ooohhh! So it would just be 9:1

    ?

    Thanks so much!
     
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