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Physics question

Discussion in 'Chit Chat' started by cowofwar, Oct 17, 2002.

  1. cowofwar

    cowofwar Peasant

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    This is basic stuff, I'm sure someone will be able to figure out what I'm doing wrong.

    The distance an object moves along a line is given by [x = at^4 - bt^2 + ct] where t is in seconds. Find the formula that describes its velocity as a function of time.

    Anways, this part is easy. Just take the derivative of the function in respect to the variable t. Which gives

    d/dt x = 4at^3 - 2bt + c

    with the values subbed in

    d/dt x = 4(0.20)t^3 - 2(0.50)t + c
    d/dt x = 0.80t^3 - t + 4.00

    Now in a previously part of the question they asked for the velocity after 11.0s. Which clearly is just the following after substitution

    d/dt x = 1.06E3m/s

    Now in this last part which is not giving me the correct answer they want the average velocity for the first 11s. So basically I'm taking the initial velocity at 0s, adding it to the final velocity at 11s and dividing it by 2 according to the law

    Vav = (Vo + V)/2

    Which gives me

    Vav = (4.00 + 1057.8)/2
    Vav = 530.9m/s

    And which of course is wrong. Someone figure out what the hell I'm doing wrong.
     
  2. cowofwar

    cowofwar Peasant

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    Or is it the area under the graph (ie: rise over run) for the function at -1 < x < 12?
     
  3. Keeksy

    Keeksy Peasant

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    Try taking the integral of v(t). Then integrate a(t) from 0 to 11.
     
  4. cowofwar

    cowofwar Peasant

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    No can do. We're not allowed to use integral calculus at the moment.
     
  5. Keeksy

    Keeksy Peasant

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    ( x(0) - x(11) ) / 11 will give you the right answer. rise over run is average velocity. the derivative of x(t) is insitaneous velocity.

    just ignore what I said earlier. I think you were on the right track.
     

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