Equation of motion with proportional drag

by Hannibal123
Tags: drag, equation, motion, proportional
 P: 19 1. The problem statement, all variables and given/known data the forces will be like this $$m⋅dv/dt=-m⋅g-k⋅v$$ I need to find the velocity equation and the place equation, thereby meaning the integral of the velocity equation. for the condition t,v=0,v_0 3. The attempt at a solution i have found this $$v(t)=mg/k*(-1+e^(-k/m⋅t))+v_0⋅e^(-k/m⋅t)$$ or in picture form http://imgur.com/WjksG But I am not sure it's correct
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P: 10,759
 Quote by Hannibal123 1. The problem statement, all variables and given/known data the forces will be like this $$m⋅dv/dt=-m⋅g-k⋅v$$ I need to find the velocity equation and the place equation, thereby meaning the integral of the velocity equation. for the condition t,v=0,v_0 3. The attempt at a solution i have found this $$v(t)=\frac{mg}{k}(-1+e^{-\frac{k}{m}t})+v_0⋅e^{-\frac{k}{m}t}$$ or in picture form http://imgur.com/WjksG But I am not sure it's correct
It looks correct, but v0=0 .

Edit: I misread vo, so it is correct.

ehild
 Homework Sci Advisor HW Helper Thanks P: 9,922 Your answer looks right to me. Why do you doubt it?
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P: 9,922
Equation of motion with proportional drag

 Quote by ehild It looks correct, but v0=0 . ehild
I read the OP condition as initially (t,v) = (0, v0)
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P: 10,759
 Quote by haruspex I read the OP condition as initially (t,v) = (0, v0)
Sorry, I misread it. Edit my post.

ehild
P: 19
 Quote by haruspex I read the OP condition as initially (t,v) = (0, v0)
that is what i meant. I doubt the results because when i draw them in a graph they dosen't seem to be correct. I have drawn them i geogebra if you are familiar with that software.
My values are:
m: 0,145 kg
k: 0,0032
g: (gravity acceleration) 9,82
v_0: 9,93 m/s

this is what i have entered in geogebra: g(x) =(0.145*9.82/0.0032)*(-1 + ℯ^(0x / 0.15)) + 9.93ℯ^(0x / 0.15)
if you want to try for yourself (should save you some time). These values are not meant for a proportional drag, but still it seems weird that the graph looks like one of a constant function
http://imgur.com/FFElN (the red one)
 P: 19 Furthermore when i integrate the velocity equation i get this http://imgur.com/IO6G6 Again not sure if I am correct
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 Quote by Hannibal123 m: 0,145 kg k: 0,0032 g: (gravity acceleration) 9,82 v_0: 9,93 m/s this is what i have entered in geogebra: g(x) =(0.145*9.82/0.0032)*(-1 + ℯ^(0x / 0.15)) + 9.93ℯ^(0x / 0.15)
How did -k/m end up as 0 / 0.15? Looks like rounding error, and the red line graph seems to be a consequence.
 P: 19 Turns out i did make some input mistakes, this looks better http://imgur.com/eVRwQ EDIT* Nope wrong again, this should be the correct equation http://imgur.com/bJ19e
 P: 19 this is the equation of place (if that is the correct term in english) as far as I am concerned the height is not suposed to be negative to positive time values??? http://imgur.com/CxAYC
 Homework Sci Advisor HW Helper Thanks P: 9,922 Your velocity time graph shows velocity +ve at time 0, so the distance time graph should show distance increasing at time 0.
 P: 19 Indeed it should, however it dosen't. Is this a wrong integration of the velocity equation? http://imgur.com/A6BKT Based on the gaph it does not seem to be correct.
 P: 19 Just found the correct place equation i think http://imgur.com/quiLu can you see a good way to shoten it?
 Homework Sci Advisor HW Helper Thanks P: 9,922 Looks right. I'd write it as $\frac{mg}{k}\left(-t+\left(\frac{m}{k}+\frac{v_0}{g}\right)\left(1-e^{-\frac{kt}{m}}\right)\right)$

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