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Integration (terminal velocity)

  1. May 29, 2007 #1
    i have 4 equations that need to be integrated

    1.) mg-kv-u=m*dv/dt

    2.) mg-kv-u=m*v(dv/ds)

    3.) mg-kv^2-u=m*dv/dt

    4.) mg-kv^2-u=m*v(dv/ds)


    **if it helps... the values are

    m=3000
    g=10
    u= 172000

    v changes so leave it as v

    any help is good help

    if those are finished can someone help me with the terminal velocity?
    i think its when t --> infinity
     
  2. jcsd
  3. May 29, 2007 #2

    cepheid

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    You know what would be really helpful? Using the homework template provided on this forum i.e.

    1. Stating the *exact* wording of the problem. (I highly doubt that "integrate these four equations" is exactly what you were asked to do. That statement doesn't even specify what the equations should be integrated with respect to! Note that these are first order differential equations for v(t), and two of them are non-linear. Solving them is likely to be more complicated than merely integrating.)

    2. *Showing what work you have done so far*. We aren't going to do your homework for you. That doesn't benefit you in the slightest. We can guide you in the right direction, so that you learn things for yourself (which is the whole point).

    That having been said, I can see that you have equations saying that the net downward force on a (presumably falling) object is equal to the downward force of gravity minus the upward drag force and also minus some other mysterious force, "u". What is u? And what exactly is the problem you have been given to solve?

    Terminal velocity is not the velocity as t --> infinity. This makes little sense, as the object is unlikely to keep falling forever. Terminal velocity occurs when the drag force becomes large enough to counteract the gravitational force. As a result, the net force on the object is then zero, and it cannot speed up any further. It continues to fall at this contant velocity...the terminal velocity.
     
    Last edited: May 29, 2007
  4. May 30, 2007 #3
    1.) mg-kv-u=m*dv/dt

    left side with respect to dv
    right side with respect to dt

    2.) mg-kv-u=m*v(dv/ds)

    left side with respect to dv
    right side with respect to ds


    3.) mg-kv^2-u=m*dv/dt

    left side with respect to dv
    right side with respect to dt

    4.) mg-kv^2-u=m*v(dv/ds)

    left side with respect to dv
    right side with respect to ds
     
  5. May 30, 2007 #4
    for question 4.) mg-kv^2-u=m*v(dv/ds)

    i got (using the constant values

    1/0.06*ln(|142000+0.03v^2|)=-1/3000*S+c

    when S=0, v=1000

    therefore c=200.921

    -1/3000*S=1/0.06*ln(|142000+0.03v^2|)-200.921

    when v=0, S=?

    therefore S=9583.88 m

    using this same method i need the others for 1.), 2.) and 3.)
     
  6. May 30, 2007 #5

    cepheid

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    I don't think you read all of my first post:

    - you still haven't used the template

    - you still haven't explained what u is

    - you still haven't stated the exact wording of the problem*

    - you still haven't really shown your work**


    *Post #3 definitely isn't the original problem, because it makes no sense for 2 reasons:

    a. One integrates with respect to v or t, NOT dv or dt. dv and dt are symbols used to denote integration with respect to v or t, respectively.

    b. MORE IMPORTANTLY, you can't integrate the two sides of an equation with respect to different variables! Didn't this set off alarm bells in your head? It makes no sense. What you do to one side of the equation, you must do to the other (if you want the equality to hold true).

    **When I say you didn't really show your work, what I mean is that you just wrote down the results of some integration operation, without putting in the steps that led to that result, making it impossible to determine what operation you carried out. (And substitution of the numbers really isn't helping in that regard. It makes it even harder to make heads or tails out of what you've done).
     
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