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Position function x(t) question

  1. Sep 13, 2010 #1
    the problem is given a(t)=4(t+3)^2, v0=-1, x0=1 find x(t), the position function

    to my knowledge x(t)= 1/2at^2+v0t+x0

    I tried it and got the wrong answer, the answer the book gives is: 1/3(t+3)^4 - 37t -26

    I did:

    x(t)= 1/2 [4(t+3)^2)]t^2 - 1t +1
    eventually getting: 4t^2 + 24t + 9 for the part in parenthesis
    and 2t^4 +12t^2 + 9/2t^2 after multiplying by t^2 and then 1/2

    What I'm getting it shaping up to look nothing like the answer, so I know I'm missing something big here.
     
  2. jcsd
  3. Sep 13, 2010 #2

    vela

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    This expression for x(t) relies on the assumption that the acceleration is constant, which is not the case here.
    By definition, you have a(t)=v'(t) and v(t)=x'(t). If you start with a(t)=constant, you can derive the equation x(t)=x0+v0t+1/2at2. This is probably covered in your textbook or notes.

    In this problem, you want to do the same thing except start with the given expression for a(t).
     
  4. Sep 13, 2010 #3
    Its for a course that already expects you to remember to remember this, we're just doing the quick review chapter in the beginning. I've never taken physics though, so I hardly every deal with this stuff, so I forgot exactly how to go about the problem. So are you saying then that I want to integrate a(t)?
     
  5. Sep 13, 2010 #4
    *integrate a(t) twice
     
  6. Sep 13, 2010 #5

    vela

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    Yes, and use the initial conditions to determine the constants of integration.
     
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