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Non-Constant Acceleration

  1. Jan 30, 2009 #1
    1. The problem statement, all variables and given/known data

    The acceleration of an object in a fluid is proportional to its speed squared. If the object's initial speed is 1.13 m/s, how much time until its speed is reduced by half (0.565 m/s)?

    2. Relevant equations

    a=-.48v^2

    3. The attempt at a solution

    I tried using the formula

    v(f)-v(0) + a*t

    Taking the derivative of the whole thing seems like it's fruitless. If I integrate the formula I can get an equation that relates to velocity:

    v=-.16x^3

    But I don't know how that relates to the time. Should the "x" be displacement or time since velocity (the result of the equation) is displacement AND time? I can solve for x, but I don't know what I'm actually solving. thanks everyone
     
  2. jcsd
  3. Jan 30, 2009 #2
    Your equation relating acceleration to velocity can be written as

    [tex]a=\frac{dv}{dt}=kv^2[/tex]

    Separate the variables and integrate.
     
  4. Jan 30, 2009 #3

    Nabeshin

    User Avatar
    Science Advisor

    Remember that these quick little equations of motion most of us memorize are all derived from the case of constant acceleration!
     
  5. Jan 30, 2009 #4
    Ahh, the acceleration IS the derivitive. So the integral with -.16 is the equation I need to use? What is the X in that equation then? time or distance?
     
  6. Jan 31, 2009 #5
    Separating variables gives

    [tex]\frac{dv}{v^2}=kdt[/tex]

    Integrate both sides using the appropriate limits for v (v0 to v0/2) and t (0 to t).
     
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