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Variable Acceleration - Clueless

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data
    The problem is labeled Variable Acceleration.

    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 the expression Ax = -3.00 Vx^2. If the marble enters this fluid with a speed of 1.50 m/s, how long will it take before the marble's speed is reduced to one-half of it's initial value.


    2. Relevant equations
    I'm not sure (I suspect that the equations of motion are used?)


    3. The attempt at a solution
    The professor has given us the answers so the answer is .222 seconds.
    I am not able to attempt a solution I can only restate the variables in more readable terms.

    Final speed (Vf) = .75
    Initial speed (V0) = 1.50
    Initial time (T0) = 0
    Initial position(X0) = 0
    Equation for var accel (Ax) = -3.00 Vx^2
    Final time (Tf) = ? this is the goal

    I know the problem has to do with integration but I'm not sure where to start.
    Can someone help me out?

    Thanks,
    Cory
     
  2. jcsd
  3. Feb 4, 2012 #2

    Curious3141

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    Have you learnt differential equations? You'll definitely need to set one up to solve this.
     
  4. Feb 4, 2012 #3
    I did differentiation and integration in calc 1 but I'm really not sure how to apply it here.
     
  5. Feb 5, 2012 #4

    SammyS

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    Hopefully you know that [itex]\displaystyle a=\frac{dv}{dt}\,.[/itex]

    Substituting (-3.00)v2 in for a gives:
    [itex]\displaystyle (-3.00)v^2=\frac{dv}{dt}\,.[/itex]​

    Divide both sides by v2 and integrate with respect to time.
     
  6. Feb 5, 2012 #5
    I am definitely missing something big here.

    I see that [itex]\displaystyle a=\frac{dv}{dt}\,.[/itex]

    and I know [itex]\displaystyle (-3.00)v^2=\frac{dv}{dt}\,.[/itex]
    because we are given [itex]\displaystyle a=(-3.00)v^2[/itex]

    As you said divide each side by v^2 (I guess so that we have all the variables on the right.
    So we get [itex]\displaystyle -3.00= \frac{dv}{(dt)(v^2)}\,[/itex]

    It can be rewritten to have dt * v^-2 on top and dt on the bottom.

    So now -3 = (dv*v^-2) / dt ... I am given initial and final velocity which I know have to go into this equation somehow. Initial = 1.5, final=.75

    I am completely lost again :(
    By virtue of knowing the answer I know that 1.5^-2 + .75^-2 ... yields the answer ... but I have no true idea why because I don't understand where that dv/dt went and I didn't really integrate (aka add one to the exponent and divide by that exponent)

    I know I am not seeing something clearly. It has been a few years since calculus but I feel like its not the physics or the calculus that are messing me up, but where they intersect.

    Thanks for your time,
    Cory
     
  7. Feb 5, 2012 #6

    SammyS

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    OK! You have [itex]\displaystyle -3.00= v^{-2}\frac{dv}{dt}\,.[/itex]

    Now integrate:
    [itex]\displaystyle \int-3.00\,dt= \int v^{-2}\frac{dv}{dt}dt[/itex]

    [itex]\displaystyle =\int v^{-2} dv [/itex]​

    Proceed on.
     
  8. Feb 5, 2012 #7
    Wouldn't the integration of -3 be -3X(or -3T in this circumstance?)

    And on the right side v^-2 should become v^(n+1)/n+1 ... Yielding (v^-1)/1 ... or v^-1 ?

    So -3T = v^-1? And plug in the final and then initial velocity?
    -3T = 1.5^-1 === -.222
    -3T = .75^-1 === -.444

    This is very confusing to me would it be -.444(equation of time with final velocity) minus -.222(equation of time with initial velocity) ... = -.222 ... Do I just disregard the negative because it is the time? Or have I completely screwed this up?

    Thanks!
    Cory
     
  9. Feb 5, 2012 #8

    Curious3141

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    The indefinite integral of [itex]v^{-2}[/itex] wrt v is [itex]-v^{-1} + C= -\frac{1}{v} + C[/itex]. You missed out the negative sign in yours (remember that the "n+1" here is MINUS one) and the constant of integration.

    I've always found it better to use definite integration on both sides and incorporate the initial conditions that way. This way you don't have to bother to solve for the constant of integration.

    So rewrite as:

    [tex]\int_0^t (-3)dt = \int_{1.5}^v v^{-2}dv[/tex]

    and here, you're basically saying that at t=0, v = 1.5 m/s (lower bound). At an unknown time t, the velocity is just v (upper bound). Although this is not the best way to write the integral from a purist mathematical perspective, it's very common in physics, and hence is acceptable.

    Now integrate and impose the bounds:

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

    See how the bounds were incorporated there. On the left hand side, the term for t=0 vanishes. On the right hand side, you need to subtract off the -1/v expression at v = 1.5, but subtracting a negative is like adding it (minus*minus = plus).

    All you're required to do now is to find t when v is half the initial velocity, or 0.75 m/s.
     
  10. Feb 5, 2012 #9
    So I was just a tad bit off for the integration (meant to put the negative in front of the v^-1)

    So once we have the basic form
    -3T = -v^-1

    Which I think I was able to get to.
    You have to add the initial velocity since that is where the equation starts?
    I'm guessing that's something you just have to think about and know to do.

    After that I plugged in .75 into v and ended up with t = .222 which is the right answer :)
    I double checked by plugging in the initial values (T=0) and (V=1.5) to make sure that both sides of the equation balanced. They did so I knew it was correct :D


    I have one last question for ya.
    The dv or dt don't really matter do they? Are they just an indicator of what we are integrating in respect to?

    Thanks so much you are a life saver. I know one of these questions is going to be on the exam (Only 8 questions on it!)

    Thanks,
    Cory
     
  11. Feb 5, 2012 #10

    Curious3141

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    It's unclear what you're doing, so I can't tell if you're really doing the right thing. I don't understand what you mean by "basic form" here. The velocity at time t is *not* given by "-3T = -v^-1" because you haven't taken into account the fact that the initial velocity is 1.5 m/s. You either have to consider the constant of integration (and then solve for it using the initial conditions), or you have to set up the definite integral as I did. Either way, what I posted is the actual relation between v and t.

    Of course they matter! They're the variables of integration. Manipulating them algebraically to bring them on either side of the equation (called separating the variables) is just a way to solve this very simple differential equation.
     
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