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Mechanical acceleration and veloctiy relationship

  1. Nov 30, 2011 #1
    1. The problem statement, all variables and given/known data

    The motion of a particle is expressed by a=42-12x2 where a is in m/s2 and X is in m, with the initial condition: v=0 when x=0 and t=0
    Determine v when X=6
    Determine X when v becomes 0 again

    2. Relevant equations
    I tried using a=dv/dt so i took the anti derivative of a to find the equation of v which turned out to be 42x-4x^3 + c and then pluged in 6 for x and found v=-612m/s

    but then for the second part when i tried finding the roots of the equation b i have two answers +3.2 and -3.2

    So i dont know if i should use +3.2 or -3.2
    Can somone tell me if im doin anything wrong? :(


    3. The attempt at a solution
     
  2. jcsd
  3. Dec 1, 2011 #2
    Note that a = [itex]\frac{dv}{dt}[/itex]

    and so v = [itex]\int[/itex]adt but v [itex]\neq[/itex][itex]\int[/itex]adx.
     
  4. Dec 1, 2011 #3
    so we would integrate (a*dt) but im not sure wat dt is :P
     
  5. Dec 1, 2011 #4
    I think that you know that the symbol

    [itex]\int[/itex]...dt

    means

    'integrate ... with respect to t'
     
  6. Dec 1, 2011 #5
    adt does not mean that a is multiplied by anything.
     
  7. Dec 1, 2011 #6
    You can't integrate your equation with respect to t as your equation doesn't rely on t (which it should in an ideal world). Maybe you copied down wrong?
     
  8. Dec 1, 2011 #7
    One can use the following:

    a = [itex]\frac{dv}{dt}[/itex] = [itex]\frac{dv}{dx}[/itex][itex]\frac{dx}{dt}[/itex] = v[itex]\frac{dv}{dx}[/itex]

    hence ∫a dx = ∫v dv = [itex]\frac{v^{2}}{2}[/itex] + constant
     
  9. Dec 2, 2011 #8
    Clever.
     
  10. Dec 2, 2011 #9
    I was reminded about that other method by a recent post on PHYSICS FORUMS!
     
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