Theoretical Mechanics: velocity dependent force

[PsychonautQQ]

Homework Statement

A projectile is launched horizontally. A retarding force acts on the projectile of the form F = -Ae^(cv). Find it's speed as function of time.

Homework Equations

F = ma

The Attempt at a Solution

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

e^(-cv) dv = (-A/m) dt

I integrated the left side between intitial velocity and final velocity, (v_i and v_f) and the right side between 0 and t. After evaluating the elementary integral and doing a small amount of algebra we come to the following:

e^(-c*v_f) = (t*A*c)/m + e^(-c*v_i)

taking the natural log of both sides...

-c*v_f = ln ((t*A*c)/m + e^(-c*v_i))

so...

v_f = (-1/c) ln((t*A*c)/m + e^(-c*v_i))

which is what I wrote for my answer. Hurray. Except an online source that is pretty dependable gives a different solution. They have the following:

v_f = v_i - (1/c)*ln(1+(A*c*t*e^(-c*v_i)/m))

Anyone want to shed some light on my mistake or tell me my other source is incorrect?

 

[TSny]

so...

v_f = (-1/c) ln((t*A*c)/m + e^(-c*v_i))

which is what I wrote for my answer. Hurray. Except an online source that is pretty dependable gives a different solution. They have the following:

v_f = v_i - (1/c)*ln(1+(A*c*t*e^(-c*v_i)/m))

Are you sure that the negative sign in their answer (shown in red) is correct?

Try the following in your answer. Factor out e^(-c*v_i) from the entire argument of your log function. Then see if you can get your answer to reduce to theirs.

 

[PsychonautQQ]

Seeing as how much log function takes the form of ln(1+ce^(-cv)) is there even a way to factor it out? Also i'll recheck my work and see if the minus sign is out of place. Did you get the same answer as me but with a positive sign? I'm pretty sure they are different answers though because they lead to different conclusions about certain aspects of the objects motion, for example how long it is traveling for. To solve for the time just said v_f equal to zero and solve for t, my equation and their equation seem to give separate results.

 

[TSny]

I think your answer is correct, but "their" answer has the wrong sign in the exponential part.

Suppose you write the argument of your ln function as (t*A*c)/m + e^-c*v_i = e^-c*v_i [e^+c*v_i(t*A*c)/m + 1]. Then use properties of the ln function to simplify.

 

[nrqed]

Seeing as how much log function takes the form of ln(1+ce^(-cv)) is there even a way to factor it out? Also i'll recheck my work and see if the minus sign is out of place. Did you get the same answer as me but with a positive sign? I'm pretty sure they are different answers though because they lead to different conclusions about certain aspects of the objects motion, for example how long it is traveling for. To solve for the time just said v_f equal to zero and solve for t, my equation and their equation seem to give separate results.

TSny is correct, there must be a sign mistake in the answer they give. With that modification, your answer is the same as theirs which can be shown using the hint given by TSny.

It is clear that you're a sneer is correct and the form they wrote is incorrect because we can look at the limit of an infinite force by taking the limit $c \rightarrow \infty$. Your answer gives v_f =0 whereas their original answer would give v_f = v_i.