Question

Newton's second law gives m (dv/dt) = mg - cv.
Show that the solution to the differential equation is v=(mg/c)(1-e^(ct/m)).
Thanks in advance!

Answer 1

Original equation:
\[m \frac{ dv }{ dt }=mg-cv\]
divided by m and rearranged:
\[\frac{ dv }{ dt }+\frac{ cv }{ m }=g\]
Integration factor:\[IF=e ^{\int\limits_{}^{}\frac{ c }{ m }dt}=e ^{\frac{ ct }{ m }}\]

Answer 2

\[\frac{ d }{ dt }(e ^{ct/m}v)=\int\limits_{}^{}g e ^{ct/m}dt\]

Answer 3

Correction:
\[\frac{d}{dt}(e^{ct/m}v)=ge^{\frac{ct}m}\]
Which then implies that:
\[e^{\frac{ct}m}v=\int \frac{d}{dt} (e^{\frac{ct}m}v) dt = \int ge^{\frac{ct}m}dt \]

Answer 4

Can you solve from there?

Answer 5

@Bobo-i-bo I'm not too sure, I can never get the correct answer

Answer 6

Can i see your attempt? Then i will be able to see where you went wrong

Answer 7

\[v e ^{ct/m}=\frac{ mg }{ c } e ^{ct/m}\]

Answer 8

you need to add an integration constant ;)

Answer 9

... +C

Answer 10

haha, so what next?

Answer 11

do the exponentials cancel?

Answer 12

You need to determine the "+C". But from the information you have given me... you can't determine C... so you need to give me more info!

Answer 13

I'm guessing the extra info needed is that v=0 when t=0? ;P

Answer 14

okay, the question was:
An object with mass m is dropped from rest and we assumed that air resistance is proportional to the speed of the object. If s(t) is the distance dropped after t seconds, then the speed is v=s'(t) and the acceleration a=v'(t). If g is the acceleration due to gravity, then the downward force on the object is mg-cv, where c is a positive constant and Newton II gives .... > that equation < ....

Answer 15

so t=0 when v=0?

Answer 16

"An object with mass m is dropped from rest" :P

Answer 17

ahhhh so the integration constant is -mg/c

Answer 18

:)

Answer 19

ah sneaky, thank you very much!