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Short calculus task

  1. Apr 12, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle initially moving at 5m/s has an acceleration of a=-.5v.
    Find v(t)


    2. Relevant equations



    3. The attempt at a solution
    i know the anti derivative of acceleration is -(v^2)/4
    but i have no idea how to put velocity in terms of time when acceleration in terms of v is given
     
  2. jcsd
  3. Apr 12, 2009 #2
    It's a separable differential equation:

    [tex]a = \frac{dv}{dt} = -\frac{1}{2}v[/tex]

    Does that help?
     
  4. Apr 13, 2009 #3
    not really...i still don't fully understand how to put v as a function of time
     
  5. Apr 13, 2009 #4
    When solving a separable differential equation, you treat the derivative (dv/dt) as a fraction. Get all the v's and dv's on one side and all the t's and dt's on the other side:
    dv/dt=-1/2 v
    1/v dv=-1/2 dt

    Then integrate both sides and solve for v.
     
  6. Apr 13, 2009 #5
    the answer given in the notebook is "v=5 exp(-.5t)"...i don't understand :s
    when i integrate i get
    ln|v| = -1/2t
    solving for v = e^(-.5t)
    where is this 5 coming from??
     
    Last edited: Apr 13, 2009
  7. Apr 13, 2009 #6
    You forgot the constant in your integration. What you have is an initial condition problem: v(0)=5. So when you integrate both sides, you get
    ln|v| = -1/2 t + c
    v=e^(-1/2 t+c)

    It makes things neater to get that constant out of the exponent, using the laws of exponents:
    v=e^c*e^(-1/2 t)

    Now, since c is a constant, e^c is a constant, so we can just cal lthat a new constant to make it neater:
    Let k=e^c
    v=k*e^(-1/2 t)

    With your initial condition v(0)=5:
    5=k*e^(-1/2*0)
    5=k*1
    k=5

    So v=5e^(-1/2 t)
     
  8. Apr 13, 2009 #7
    ahh i c thx alot!
     
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