Velocity dependent force question

[PsychonautQQ]

Homework Statement

Initial speed = v, only for acting on it is retarding force F(v) = -Ae^(-cv). Find it's speed as a function of time.

Homework Equations

F=ma

The Attempt at a Solution

m(dv/dt) = -Ae^(-cv)
dv / (-Ae^(-cv) = dt / m

integrating gives
[e^(-cv)/Ac] = t/m

where the left side of that integral is evaluated from initial vi to final vf

e^(-cvi)/Ac-e^(-cvf)/Ac = t / m

multiplying by AC and taking the natural log of all of this...

-cvf + cvi = ln(tAc/m)
vf = vi - (1/c)*ln(tAc/m)

Does this look correct? An online source says this is wrong. Thanks for the help, LaTex coming soon.

 

[vanhees71]

Well, check your signs. Principally, using "separation of variables" is the correct approach to solve this 1st order ODE.

 

[HallsofIvy]

Homework Statement

Initial speed = v, only for acting on it is retarding force F(v) = -Ae^(-cv). Find it's speed as a function of time.

Homework Equations

F=ma

The Attempt at a Solution

m(dv/dt) = -Ae^(-cv)
dv / (-Ae^(-cv) = dt / m

integrating gives
[e^(-cv)/Ac] = t/m

You do know that $1/e^{-cv}= e^{cv}$, don't you? So simpler is
$$me^{cv}dv= -Adt$$
and the integrating
$$\frac{m}{c}e^{cv}= -At+ C$$

where the left side of that integral is evaluated from initial vi to final vf

e^(-cvi)/Ac-e^(-cvf)/Ac = t / m

multiplying by AC and taking the natural log of all of this...

-cvf + cvi = ln(tAc/m)
vf = vi - (1/c)*ln(tAc/m)

Does this look correct? An online source says this is wrong. Thanks for the help, LaTex coming soon.