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Math Help - Motion in a resisting medium

  1. #1
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    Smile Motion in a resisting medium

    Hello everyone.. I'm having trouble with solving this diff equation

    m. dv/dt = mg - k.v^2

    Show that the solution to this differential equation is,

    v = (gm/k)^1/2 x (e^ct -1 / e^ct + 1), where c = 2(kg/m)^1/2


    Please help me^^
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  2. #2
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    Quote Originally Posted by Modernized View Post
    Hello everyone.. I'm having trouble with solving this diff equation

    m. dv/dt = mg - k.v^2

    Show that the solution to this differential equation is,

    v = (gm/k)^1/2 x (e^ct -1 / e^ct + 1), where c = 2(kg/m)^1/2


    Please help me^^
    What have you tried? Did you at least get this far: \frac{dt}{dv} = \frac{m}{mg - k v^2} ?

    What do you think the next step is?

    Please show the working you've done and state where you get stuck.
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  3. #3
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    Smile

    Hi! Thank you so much for replying.
    This is what I've got so far,
    m. dv/dt = mg - kv^2
    dv/dt = g - k/m.v^2
    dv/dt = -k/m ( -mg/k + v^2)

    Bring it to the other side,
    1 / (-mg/k + v^2) .dv = -k/m dt
    Integrate both sides, however we need to use table of integrals here.
    (1/2a ln | a+x / a-x | + c)
    So, I got
    1 / 2(mg/k)^1/2 ln | (mg/k)^1/2 +v / (mg/k)^1/2 -v | = -kt/m + c

    This is what I got so far.. Im not sure how to go on from here. Thank you so much
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    [snip]
    \frac{dt}{dv} = \frac{m}{mg - k v^2} ?

    [snip]
    \Rightarrow t = \sqrt{\frac{m}{gk}} \ln \left( \frac{ \sqrt{mg} + \sqrt{k} v }{\sqrt{mg} - \sqrt{k} v }\right) + C


    \Rightarrow A e^{\sqrt{\frac{m}{gk}} t} = \frac{ \sqrt{mg} + \sqrt{k} v }{\sqrt{mg} - \sqrt{k} v }

    Is there a given boundary condition (so that A can be found)?
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  5. #5
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    yes!
    A skydiver has a terminal velocity of 50m/s before opening her parachute, g = 9.8m/s^2 and her mass is 60kg, find k and her speed after 2 and 6 seconds of free-fall.

    By the way, I understand how in your first step you fliped it to make t the subject.

    But.. is my way of doing it correct? Or do I have to do it like your way?..
    when I find t as the subject.. isn't it harder to find v as the subject at the end?
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  6. #6
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    Quote Originally Posted by Modernized View Post
    yes!
    A skydiver has a terminal velocity of 50m/s before opening her parachute, g = 9.8m/s^2 and her mass is 60kg, find k and her speed after 2 and 6 seconds of free-fall.

    By the way, I understand how in your first step you fliped it to make t the subject.

    But.. is my way of doing it correct? Or do I have to do it like your way?..
    when I find t as the subject.. isn't it harder to find v as the subject at the end?
    If your work is correct it will reduce to what I have done.

    Were you told that v = 0 when t = 0? If not, you're probably meant to assume it. Use it to get the constant of integration. Then re-arrange the expression to make v the subject.

    Since the terminal velocity is 50 m/s and you know that at terminal velocity acceleration is equal to zero, you can say 0 = mg - k(50)^2 ....
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