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Acceleration in fluid

  1. Sep 22, 2004 #1
    Question states:

    The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared and is given (in SI units) by a=-3.00v^2 for v>0. If the marble enters this fluid with a speed of 1.50m/s, how long will it take before the marble's speed is reduced to half of its initial value?

    Here's what I think is it...take the derivative of a... a`=6v

    d=6(.75m/s)
    d=4.5m

    I need the acceleration to sub in for the equation vf = vi + at
    to find the time.

    Am I doing any of this correct?
     
  2. jcsd
  3. Sep 22, 2004 #2
    Can't use that formula because this is not uniformily accelerated motion (acceleration is a function of velocity here). Try taking the integral of acceleration to get a v-t relationship.
     
  4. Sep 22, 2004 #3
    remember from our last conversation that I dont know how to take integrals? :) Can you teach me with an example and explain what and integral does?
     
  5. Sep 22, 2004 #4
    I could, but you should try reading about it on the internet, or better yet, in a calculus textbook, which will contain all the information you need. On a basic level, integration is a way to find areas under and between curves using only their equation. If you are not required to know calculus for your course, however, the questions are most likely doable without integration.

    Sorry, I can't explain much more now. Gotta run.
     
  6. Sep 23, 2004 #5

    Tide

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    UrbanX,

    Since you didn't specify the mass of the marble I take it that you are ignoring gravity.

    In that case
    [tex]\frac {dv}{dt} = -3v^2[/tex]
    which amounts to integrating
    [tex]\frac {dv}{v^2} = -3 dt[/tex]
    You should be able to handle it from there.
     
  7. Sep 23, 2004 #6
    I actually have no clue what happened and how to even begin. Thank you for the example but I do not know what you are doing since I do not know how to integrate
     
    Last edited: Sep 23, 2004
  8. Sep 24, 2004 #7

    Pyrrhus

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    Taking from Tide's answer. Upon request by Urban.

    [tex]\frac {dv}{v^2} = -3 dt[/tex]

    [tex]\int^{v}_{v_{o}} \frac {dv}{v^2} = \int^{t}_{0} -3 dt[/tex]

    [tex] -\frac{1}{v}]^{v}_{v_{o}} = -3t]^{t}_{0}[/tex]


    [tex] -\frac{1}{v} + \frac{1}{1.5}= -3t - 0[/tex]

    Urban, you know initial speed is 1.5, and v will be equal 1.5/2, so just solve for t.
     
    Last edited: Sep 24, 2004
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