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Finding the position x(t) with constantly changing acceleration

  1. Nov 9, 2008 #1
    Im trying to find the position of an object x at time t. There is pretty straight forward formulas to use to find x(t), however acceleration must be fixed... but in this question acceleration is changing at sin2t.

    I have been going through period by period working out the new position at x(t-1).. until i finally get to the time i want.

    surely there is an easier way to work out the position at time t when acceleration is constantly changing?
     
  2. jcsd
  3. Nov 9, 2008 #2

    Hootenanny

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    There is indeed a far more straightforward method to determine x(t), simply consider the definition for acceleration.

    [tex]a = x^{\prime\prime}\left(t\right) = \sin\left(2t\right)[/tex]
     
  4. Nov 9, 2008 #3
    thanks, i can see how that is true. However i cant put it all together.

    The object is moving at v = -1/2 at time 0, from a starting point of 5 on a grid.. how can i find out where the object is 5 seconds later?

    Is it the integral of the integral of acceleration (sin2t) where t = 5?
     
    Last edited: Nov 9, 2008
  5. Nov 9, 2008 #4

    Hootenanny

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    No, try taking it one step at a time. First determine v(t)

    [tex]v\left(t\right) = x^\prime\left(t\right) = \int \sin\left(2t\right) dt[/tex]
     
  6. Nov 9, 2008 #5
    (-1/2)cos(2t)
     
  7. Nov 9, 2008 #6

    Hootenanny

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    Good, but aren't you forgetting something?
     
  8. Nov 9, 2008 #7
    woops...

    (-1/2)cos(2t) + C
     
  9. Nov 9, 2008 #8

    Hootenanny

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    Better. So what can you say about this constant C?
     
  10. Nov 9, 2008 #9
    C = acceleration at t?
     
  11. Nov 9, 2008 #10

    Hootenanny

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    No. We have found an equation for the velocity at time t,

    [tex]v\left(t\right) = C - \frac{1}{2}\cos\left(2t\right)[/tex]

    Let me rephrase my question: what do the initial conditions tell you about the value of C?
     
  12. Nov 9, 2008 #11
    i think that C is the initial velocity of -1/2?
     
  13. Nov 9, 2008 #12

    Hootenanny

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    Close, but not quite. Try substituting t=0 into the equation for v(t) above.
     
  14. Nov 10, 2008 #13
    ok, so using C = -1/2 we get v(0) = -1, and using C = 1/2 we get v(0) = 0

    neither of which gets our true initial velocity of -1/2... the only way i found to get the correct initial velocity at t(0) is using C = 0.
     
  15. Nov 10, 2008 #14
    when t = 0 you get

    [tex]v\left(0 \right) + \frac{1}{2}= C [/tex]

    You need to look at your question to find out what the value of the velocity is at t = 0
    Once you get that, you'll find a number for C and the velocity is known as a function of time.

    Then, you can do the entire same thing :

    integrate the formula you found for v(t) to get to x(t)

    marlon
     
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