Question

\[m\frac{\text dv}{\text dt}=mg-kv;\qquad v(0)=0\]

Answer 1

\[\frac{\text dv}{\text dt}+\frac kmv=g\]\[\qquad\qquad\qquad \text{let}\frac km=\omega_0^2\]\[\frac{\text dv}{\text dt}+\omega_0^2v=g\]

Answer 2

\[\mu(t)=e^{\int \omega_0^2\text dv}=e^{\omega_0^2 v}\]

\[\frac{\text d }{\text dt}\left(v\cdot e^{\omega_0^2t}\right)=g\cdot e^{\omega_0^2 t}\]

Answer 3

\[v\cdot e^{\omega_0^2 t}=\int g\cdot e^{\omega_0^2 t}\text dt\]\[v\cdot e^{\omega_0^2 t}=ge^{\omega_0^2 }\int e^{t}\text dt\]\[v\cdot e^{\omega_0^2 t}=ge^{\omega_0^2}\left(t{e^t}+c\right)\]\[v(t)=g\left(t+ce^{-t}\right)\]\[v(0)=gc=0\]\[c=0\]\[v(t)=gt\]

Answer 4

is this correct?

Answer 5

between lines 1 and 2 on reply 3, I don't think you can factor out omega like you did

Answer 6

\[e^n * e^m = e^{n+m}\]

Answer 7

\[\int\limits e^{t}\text dt\]

= ??

Answer 8

e^t

Answer 9

i knew i was doing something wrong in the middle section

Answer 10

Then how did t come with this in third step ??

Answer 11

In fourth step sorry..

\[v\cdot e^{\omega_0^2 t}=ge^{\omega_0^2 }\int e^{t}\text dt\]

After this ??

Answer 12

I get:
\[\frac{gm}{k} + C\]

Answer 13

\[\frac{\text dv}{\text dt}+\omega_0^2v=g\]\[\mu(t)=e^{\int \omega_0^2\text dt}=e^{\omega_0^2 t}\]\[\frac{\text d }{\text dt}\left(v\cdot e^{\omega_0^2t}\right)=g\cdot e^{\omega_0^2 t}\]

Answer 14

Can't find anything wrong with that

Answer 15

Then again it is 3:30 AM where I live

Answer 16

\[v\cdot e^{\omega_0^2 t}=\int\limits_0^t g\cdot e^{\omega_0^2 t'}\text dt'\]\[v\cdot e^{\omega_0^2 t}=ge^{\omega_0^2 }\int\limits_0^t e^{t'}\text dt'\]\[v\cdot e^{\omega_0^2 t}=ge^{\omega_0^2}\left(e^t+1\right)\]

Answer 17

is this better ?
\[v(t)=g\left(1+ce^{-t}\right)\]\[v(0)=g(1+c)=0\]\[c=0\]\[v(t)=gt\]

Answer 18

I don't think you can factor a:
\[e^\omega \]
out of the integral

Answer 19

You can take out uncle rocks as that is the constant only..

Answer 20

\(m\) and \(k\) are constants

Answer 21

Yes so omega will also be constant..

Answer 22

but:
\[e^{\omega t} \neq e^ \omega * e^t\]

Answer 23

and in fact if you evaluate the two integrals you will see that they are different

Answer 24

by a factor of omega

Answer 25

or whatever we are using, omega^2

Answer 26

Did not you get c= -1 instead of 0 @UnkleRhaukus

Answer 27

hmmmmmmm

Answer 28

im a little confused

Answer 29

\[\large v\cdot e^{\omega_0^2 t}=\int\limits\limits_0^t g\cdot e^{\omega_0^2 t'}\text dt'\]

\[\large v\cdot e^{\omega_0^2 t} = g \left| \frac{e^{(\omega_0)^2t'}}{(\omega_0)^2} \right|_{0}^{t}\]

Answer 30

\[v\cdot e^{\omega_0^2 t}=\int\limits_0^t g\cdot e^{\omega_0^2 t'}\text dt'\]\[v\cdot e^{\omega_0^2 t}=g\int\limits_0^t e^{\omega_0^2 t'}\text dt'\]\[v\cdot e^{\omega_0^2 t}=g\left.\frac {e^{\omega_0^2t'}}{\omega_0^2}\right|_0^t \]\[v\cdot e^{\omega_0^2 t}=g\frac {e^{\omega_0^2t}-1}{\omega_0^2} \]\[v(t)=g\left(1+e^{-t}\right)\]

Answer 31

Sorry you cannot separate omega squared and t because they are in multiplication and not addition..

Answer 32

\[\large v(t)=\frac{g}{(\omega_0)^2}\left(1-e^{-(\omega_0)^2t}\right)\]

Answer 33

yes

Answer 34

\[v\cdot e^{\omega_0^2 t}=g\frac {e^{\omega_0^2t}-1}{\omega_0^2} \]\[v(t)=\frac{g}{\omega_0^2}\left(1-e^{-\omega_0^2t}\right)\]\[v(0)=\frac{g}{\omega_0^2}\left(1-1\right)=0\]\[v(t)={g}\frac{\left(1-e^{-\omega_0^2t}\right)}{\omega_0^2}\]

Answer 35

Yes now put the conditions..

Answer 36

i though i did that?

Answer 37

it seemed like a "tutorial" :)

Answer 38

well distance traveled is
\[ve^{\int\limits_{}^{}kt/m}=\int\limits_{}^{}{g}e^{\int\limits_{}^{}kt/m}dt+c\]

Answer 39

have i finished this question or what//?

Answer 40

so\[{g{e}^{kt/m}\over{{k/m}}}+c = {{gm}\over{k}}{e}^{kt/m} +c\]
so v
\[v={{gm}\over{k}}+ce^{-kt/m} \]
intial cond.
\[0=v={{gm}\over{k}}+ce^{-kt/m} \]

Answer 41

sorry my computer bugs out whenever i use the equation thing

Answer 42

try to type that in notepad and then copy paste it here @noah_ochoa :)

Answer 43

if x = distance traveled in t
and \[c=-gm/t\]

Answer 44

\[v(t)=\frac {mg}k\left(1-e^{-\frac {kt}m}\right)\]