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Acceleration, velcocity, and position

  1. Jan 7, 2010 #1
    1. The problem statement, all variables and given/known data
    The acceleration of a particle is directl porportional to the square of the time t. When t=0s, and the particle is at x=24m. Knowing that at t=6s, x=96m, and V=18m/s, express x and v in terms of t.

    2. Relevant equations


    3. The attempt at a solution

    so i took the integral of a=t2 and got v=1/3t3+C, then took the integral again to obtain x=1/12t4+Ct+A.

    So then I plugged in x=24 and t=0 to get A=24

    so now my eqation is.....
    x=1/12t4+Ct+24 solving for C i get C=-6

    so then my equation is x=1/12t4-6t+24 in the book the answer is 1/108t4+10t+24

    what am I doing wrong?
  2. jcsd
  3. Jan 7, 2010 #2


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    Homework Helper

    Directly proportional does not mean equal. So a = kt2, where k is a unknown constant. You get it from the data given for t=0 and t=6 s.
  4. Jan 7, 2010 #3
    oh, that makes more sense! Thank you!
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