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Integrating velocity function

  1. Nov 18, 2012 #1
    1. The problem statement, all variables and given/known data

    if [itex]v = 8t - 3t^2[/itex] and the body is at P

    find where the body is from P when t=3

    3. The attempt at a solution

    I am very new to calculus and have just been taught basics of differentiation and integration.

    I know that when you integrate there is an arbitrary constant so I have go this far (below).

    [itex]∫ ( v = 8t - 3t^2 ) dt = 4t^2 - t^3 + C[/itex]

    I am not sure what to do with the constant, is it ignored and thus the answer is as below

    [itex]s = 4t^2 - t^3[/itex]
    [itex]s = (4 \times 9) - 27[/itex]
    [itex]s = 36 - 27[/itex]
    [itex]s = 9[/itex]metres ?

    Or do I have to do something with the constant?
     
    Last edited: Nov 18, 2012
  2. jcsd
  3. Nov 18, 2012 #2

    rock.freak667

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    Normally at t=0, s=0 and v=0. So your constant C will work out to be 0. So what you did is correct.
     
  4. Nov 18, 2012 #3

    Ray Vickson

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    The distance travelled is s(t) - s(0), from which C drops out.

    RGV
     
  5. Nov 19, 2012 #4
    OK thanks for the help :)
     
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