differential equation problem

Question

anonymous · Answer

attachment

anonymous · Answer

i've reached to the point where $$\int\limits_{}^{}e ^{\frac{ c }{ m }t}gdt= \frac{ mg }{ c }e ^{\frac{ c }{ m }t} +c$$

anonymous · Answer

how to find v?

ganeshie8 · Answer

$\large \mathbb{ m \frac{dv}{dt} = mg -cv} $

ganeshie8 · Answer

m, g, c are constants... 
so its simply a separable equation

ganeshie8 · Answer

$\large \mathbb{ m \frac{dv}{dt} = mg -cv} $

$\large \mathbb{ m \frac{dv}{mg-cv} = dt} $

ganeshie8 · Answer

integrate  both sides ?

ganeshie8 · Answer

$\large \mathbb{ \int m \frac{dv}{mg-cv} = \int dt} $

anonymous · Answer

-(mln(gm-cv))/c = t + c?

ganeshie8 · Answer

looks good, u can plug the integration constant = 0
cuz, when t=0, velocity of object=0
(the object started from rest)

ganeshie8 · Answer

$\large \mathbb{ m \frac{dv}{dt} = mg -cv} $

$\large \mathbb{ m \frac{dv}{mg-cv} = dt} $

intgrate both sides

$\large \mathbb{ \int m \frac{dv}{mg-cv} = \int dt} $

$\large \mathbb{ -\frac{m}{c} \ln |mg-cv| = t + c_1} $

$c_1=0$,  cuz v=0, when t= 0

$\large \mathbb{ -\frac{m}{c} \ln |mg-cv| = t } $

anonymous · Answer

okay, then i just hav to make v the subject?

ganeshie8 · Answer

yup

anonymous · Answer

uhm can i check with you. $$lne ^{mg-cv}=lne^{\frac{ -ct }{ m }}$$ is this correct?

anonymous · Answer

mg-cv=-ct/m ? or the ln proprty is incorrect?

ganeshie8 · Answer

$\large \mathbb{ m \frac{dv}{dt} = mg -cv} $

$\large \mathbb{ m \frac{dv}{mg-cv} = dt} $

intgrate both sides


$\large \mathbb{ \int m \frac{dv}{mg-cv} = \int dt} $

$\large \mathbb{ -\frac{m}{c} \ln |mg-cv| = t + c_1} $

$c_1=0$ cuz v=0, when t= 0

$\large \mathbb{ -\frac{m}{c} \ln |mg-cv| = t } $

$\large \mathbb{  \ln |mg-cv| = -ct/m } $

take exponent both sides :

$\large \mathbb{  e^{\ln |mg-cv|} = e^{-ct/m }} $

ganeshie8 · Answer

$\large \mathbb{  mg-cv = e^{-ct/m }} $

solve v

anonymous · Answer

haha okay! uhm how do u know |mg-cv| is positive?

ganeshie8 · Answer

good question :)
mg > cv

so, mg-cv is positive

ganeshie8 · Answer

remember, cv is the air resistance force, 
which is always less than the gravitaional force... 

otherwise, the object will FALL UP. which is illegal

ganeshie8 · Answer

so, mg is always greater than cv
giving u mg-cv > 0

anonymous · Answer

ooo that makes sense haha, okay i shd be good thank you again! :D

ganeshie8 · Answer

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