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Particle motion question

  1. Sep 24, 2006 #1

    okay...i labeled acceleration 0 -19 as c.

    a=c
    v=cx
    d=0.5cx^2

    for 19-23 we have

    a=-9.8
    v=-9.8x
    c=-4.9x^2


    0 -19 we have d=180.5c
    19-23
    we have -823.5

    180.5c-823.5=4700

    Am I doing it right in solving for C which is constant acceleration in 0 -19?

    edit: ok I know now it's not right.

    my final distance formula would be

    4700=-4.9x^2(for 19</x</23)+0.5cx^2(0</x</19)

    But i dont know how to put this in a real equation form...
     
    Last edited: Sep 24, 2006
  2. jcsd
  3. Sep 24, 2006 #2
    my second approach...

    v=cx-9.8x

    v=19c-9.8(4)=19c-39.2

    and now i take the integral of all this and set it equal to 4700

    okay...so i think the following formula should work...

    [tex]\int_{0}^{19} {19c} dx + \int_{19}^{23} {-39.2} dx = 4700[/tex]

    grrrr. nope. Also sig figs matter.
    I got a 13.5 answer...and a 13.2 answer...this is getting fustrating.
     
    Last edited: Sep 24, 2006
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