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Homework Help: Derive y as function of time in relation to terminal velocity

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Integrate

    [tex]v_{y} = v_{t} (1 - e^{-(\frac{k}{m})t}))[/tex]

    to get [tex]y[/tex] as a function of time.

    2. Relevant equations
    [tex]y = v_{t}(t - \frac{m}{k} (1 - e^{-(\frac{k}{m})t})))[/tex] is the solution.

    [tex]v_{t}=\frac{mg}{k}[/tex]

    3. The attempt at a solution
    Well, when I integrate I get the following:

    [tex]y = v_{t}(t + \frac{m}{k} e^{-(\frac{k}{m})t})))[/tex]

    I don't get where that middle part comes from, which when taken out of brackets equals

    [tex]-\frac {v_{t}m}{k}[/tex].

    Does anyone have any idea what I'm missing or where I may have gone wrong with my integration technique?
     
    Last edited: Sep 18, 2010
  2. jcsd
  3. Sep 18, 2010 #2

    kuruman

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    No, it does not. What do you get when you distribute vT correctly inside the parentheses?
     
  4. Sep 18, 2010 #3
    I think that middle part does actually match what I wrote. I didn't mean for the whole thing to be taken out of brackets, just that thing that my solution is missing when compared to the proper one. That's what I meant by "middle part".
     
  5. Sep 18, 2010 #4

    kuruman

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    Look at the equation for y(t). When t is zero, you get correctly y = 0. When t becomes very large, you get (approximately) y = vTt, i.e. the object is moving at constant terminal velocity. So the middle part is needed to give you the correct result for the in-between motion.
     
  6. Sep 18, 2010 #5
    So this is a constant "C" that you need to add yourself after integrating? And can you explain in more detail why you need to add specifically that expression? I thought since starting velocity is zero everything (everything, but not more than that) is covered when you integrate that original velocity equation. Why isn't this the case then?
     
  7. Sep 18, 2010 #6

    kuruman

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    There is no constant of integration C. Show me the integral that you used to get the position y(t) from the velocity v(t), and I will explain why.
     
  8. Sep 18, 2010 #7
    I used this one:

    [tex]\int v_{t} (1 - e^{-(\frac{k}{m})t})) = v_{t}t - v_{t} \int e^{-(\frac{k}{m})t} = v_{t}(t + \frac{m}{k} e^{-(\frac{k}{m})t})))[/tex]
     
  9. Sep 18, 2010 #8

    kuruman

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    First off this is only the right side of the equation that you should be using. Show and integrate both sides. Secondly, what are your limits of integration?
     
  10. Sep 18, 2010 #9
    Yeah, sorry, I left out the left side, because it's just

    [tex]\int^t_0 v_{y} dt = y[/tex].

    And the limits would be [tex]\int^t_0[/tex].
     
  11. Sep 18, 2010 #10
    Facepalm.jpg would be in order here. I just forgot to apply the limits and derived an indefinite instead of a definite integral, didn't I? I'm screaming inside like a madman right now.
     
  12. Sep 18, 2010 #11

    kuruman

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    You also need limits for the y-dot integral which should be y0 to y.
     
  13. Sep 18, 2010 #12
    Thanks a lot for all of your help. One last question, though. When you mention you must have limits from y0 to y, do you actually mean from 0 to t, which when integrated return values of y - y0?
     
  14. Sep 18, 2010 #13

    kuruman

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    No. On the left side the dy integral goes from y0 to y and on the right side the dt integral goes from zero to t. The lower limit on y matches the lower limit on t and the upper limits are also matched likewise in pairs. That's why I said "Show and integrate both sides." You start with

    dy = v dt

    and you integrate both sides with lower and upper limits of integration on both sides.
     
  15. Sep 18, 2010 #14
    Hmm, but on the left-hand side you start out with vy, which you have to integrate in relation to time. So setting that up you get

    [tex]\int v_{y} dt[/tex]

    on the left. How would you then implement the limit for displacement (y and y0) if the variable is time?
     
  16. Sep 19, 2010 #15

    kuruman

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    Look

    [tex]dy=vdt[/tex]

    As time increases, add all increments dy on the left which means adding all increments v dt on the right.

    [tex]\int^{y}_{y_0}dy=\int^{t}_{0} vdt[/tex]
     
  17. Sep 19, 2010 #16
    Yeah, I understand that, but does that have any practical implications or is it just for notation purposes? I mean, in our case y0 = 0, so I'm guessing you would only need to take that into account if it wasn't. But how would you do that then? Subtract y0 from the result you would get on the right hand side?
     
  18. Sep 19, 2010 #17

    kuruman

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    What do you get when you evaluate

    [tex]\int^{y}_{y_0}dy \:?[/tex]

    That's how you do it.

    :rolleyes:
     
  19. Sep 19, 2010 #18
    I get y - y0, right?
     
  20. Sep 20, 2010 #19

    kuruman

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    Right and that's the left side of the equation. If y0 is zero in the particular situation you are dealing with, then you put nothing on the left side; if it is not zero, then you put the negative of whatever the problem tells you it is at time t=0. It all depends on the initial conditions of your specific problem. What more is there to say?
     
  21. Sep 20, 2010 #20
    Yeah, I guess I put that y0 = 0 as granted, as it was stated so in the book in the paragraphs preceding the equation, so that's why I didn't really address it at first. But once again, thanks again for your help.
     
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