
#1
Dec1912, 02:47 PM

P: 19

1. The problem statement, all variables and given/known data
the forces will be like this [tex]m⋅dv/dt=m⋅gk⋅v[/tex] 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 [tex]v(t)=mg/k*(1+e^(k/m⋅t))+v_0⋅e^(k/m⋅t)[/tex] or in picture form http://imgur.com/WjksG But I am not sure it's correct 



#2
Dec1912, 03:39 PM

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Edit: I misread vo, so it is correct. ehild 



#3
Dec1912, 03:43 PM

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P: 9,143

Your answer looks right to me. Why do you doubt it?




#4
Dec1912, 03:44 PM

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Equation of motion with proportional drag 



#5
Dec1912, 03:53 PM

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ehild 



#6
Dec1912, 04:40 PM

P: 19

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) 



#7
Dec1912, 04:42 PM

P: 19

Furthermore when i integrate the velocity equation i get this http://imgur.com/IO6G6
Again not sure if I am correct 



#8
Dec1912, 07:44 PM

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#9
Dec1912, 07:59 PM

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 



#10
Dec1912, 08:09 PM

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 



#11
Dec1912, 09:54 PM

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Your velocity time graph shows velocity +ve at time 0, so the distance time graph should show distance increasing at time 0.




#12
Dec2012, 05:29 AM

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. 



#13
Dec2012, 09:14 AM

P: 19

Just found the correct place equation i think
http://imgur.com/quiLu can you see a good way to shoten it? 



#14
Dec2012, 03:21 PM

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Looks right. I'd write it as
[itex]\frac{mg}{k}\left(t+\left(\frac{m}{k}+\frac{v_0}{g}\right)\left(1e^{\frac{kt}{m}}\right)\right)[/itex] 


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