# Finding v(t) for a velocity dependent force

1. Feb 6, 2009

### littlehonda

1. The problem statement, all variables and given/known data

Given the one-dimensional retarding force F=-Ae^(-$$\alpha$$v) find an expression for v(t).

2. Relevant equations

F = m(dV/dt)
A and $$\alpha$$ are constants, v is instantaneous speed.

3. The attempt at a solution

Im not sure how to frame the idea of integrating a velocity dependent force. Can I take simply integrate and use

v= -Ae^(-$$\alpha$$v)t

or should the v be eliminated from the right hand side. I tried doing that by dividing the left by e^(-$$\alpha$$v) and then doing a e^u du type integration but the answer i came up with didn't seem reasonable.

just looking for a little hint or push here, thanks!

2. Feb 6, 2009

### gabbagabbahey

Hint: You have a separable differential equation for v(t)

3. Feb 6, 2009

### littlehonda

thanks for responding gabba,

I think that I tried the separable differential equations method but couldnt make sense of the answer

what i did was

e^($$\alpha$$v) dv = -Adt

1/$$\alpha$$(e^($$\alpha$$v)-e^($$\alpha$$v0)) = -At

i'm not sure what this means though or where to go from here

4. Feb 6, 2009

### gabbagabbahey

$$\int e^{\alpha v}dv =\frac{e^{\alpha v}}{\alpha}+C_1$$

And

$$\int-Adt=-At+C_2$$

$$\implies \frac{e^{\alpha v}}{\alpha}=-At+C_3$$ where $C_3 \equiv C_1-C_2$

Multiply both sides of the equation by $\alpha$ and then take the natural log. What initial conditions are you given?

5. Feb 6, 2009

### littlehonda

ok so thats where I got to pretty much, the C1+C2=C3 makes sense.

Here is all the information given

A= 1/m/s^2
$$\alpha$$=.1 s/m
v0= 20 m/s

6. Feb 6, 2009

### gabbagabbahey

I assume $v_0=v(t=0)$? If so, you can solve for $C_3$.

7. Feb 6, 2009

### HallsofIvy

Staff Emeritus
$$m\frac{dv}{dt}= -Ae^{-\alpha t}[/itex] which separates as [tex]e^{\alpha v}dv= -A/m dt$$
(you dropped the "m")
Integrating,
$$-\frac{1}{\alpha}e^{\alpha v}+ C= -(A/m)t$$
Assuming that your $v_0$ is the velocity when t= 0,
$$-\frac{1}{\alpha}e^{\alpha v_0}+ C= 0$$
so
$$C= \frac{1}{\alpha}e^{\alpha v_0}$$
so
$$-\frac{1}{\alpha}\left(e^{\alpha v}- e^{\alpha v_0}\right)= -(A/m)t$$
which is just what you have (except for the "m").

Are you concerned with solving for v?
$$e^{\alpha v}- e^{\alpha v_0}= (A\alpha/m)t$$
$$e^{\alpha v}= (A\alpha/m)t+ e^{\alpha v_0}$$
Now take the logarithm of both sides.

Last edited: Feb 6, 2009
8. Feb 6, 2009

### littlehonda

$v_0=v(t=0)$

(e^20$\alpha$)/$\alpha$ = C3

C3 = 73.8

v(t) = (ln(-At$\alpha$)+ ln(73.8$\alpha$))/$$alpha$$

sorry if im a little slow with this, am I making an error here?

9. Feb 6, 2009

### gabbagabbahey

Yes, you are making a few errors.

To start with, Halls pointed out that you dropped the mass of the object from the differential equation....when you include it, you should get

$$\frac{me^{\alpha v}}{\alpha}=-At+C_3$$

Multiply both sides of the equation by alpha/m:

$$\implies e^{\alpha v}=\frac{\alpha}{m}(-At+C_3)$$

Then take the natural log of both sides

$$\alpha v =\ln\left(\frac{\alpha}{m}(-At+C_3)\right)$$

You can take it from there...

10. Feb 6, 2009

### littlehonda

I was careless putting the force equation up, the right side of the equation also contains a m to cancel the one on the left.

11. Feb 6, 2009

### littlehonda

v= (ln($\alpha$(-At+C3)))/$\alpha$

Solving for C3

C3 = (e^v0$\alpha$)/$\alpha$

at V(0) t=0

C3 = 73.89

v= (ln($\alpha$(-At+73.89)))/$\alpha$

12. Feb 6, 2009

### gabbagabbahey

Looks good to me

13. Feb 6, 2009

### littlehonda

hey thanks alot both of you guys!

hate to keep pestering but if I want to find x(t) I just integrate dx/dt and find a new constant C3 using the initial condition x(0) t=0?