Finding v(t) for a velocity dependent force

[littlehonda]

Homework Statement

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

Homework Equations

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

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^(-\alphav)t

or should the v be eliminated from the right hand side. I tried doing that by dividing the left by e^(-\alphav) 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!

 

[gabbagabbahey]

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

 

[littlehonda]

thanks for responding gabba,

I think that I tried the separable differential equations method but couldn't make sense of the answer

what i did was

e^(\alphav) dv = -Adt

1/\alpha(e^(\alphav)-e^(\alphav0)) = -At

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

 

[gabbagabbahey]

thanks for responding gabba,

I think that I tried the separable differential equations method but couldn't make sense of the answer

what i did was

e^(\alphav) dv = -Adt

1/\alpha(e^(\alphav)-e^(\alphav0)) = -At

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

\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?

 

[littlehonda]

ok so that's 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

 

[gabbagabbahey]

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

 

[HallsofIvy]

Homework Statement

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

Homework Equations

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

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^(-\alphav)t

or should the v be eliminated from the right hand side. I tried doing that by dividing the left by e^(-\alphav) 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!

thanks for responding gabba,

I think that I tried the separable differential equations method but couldn't make sense of the answer

what i did was

e^(\alphav) dv = -Adt

1/\alpha(e^(\alphav)-e^(\alphav0)) = -At

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

Your original equation is
m\frac{dv}{dt}= -Ae^{-\alpha t}[/itex]<br /> which separates as <br /> e^{\alpha v}dv= -A/m dt<br /> (you dropped the &quot;m&quot;)<br /> Integrating,<br /> -\frac{1}{\alpha}e^{\alpha v}+ C= -(A/m)t<br /> Assuming that your v_0 is the velocity when t= 0, <br /> -\frac{1}{\alpha}e^{\alpha v_0}+ C= 0 <br /> so <br /> C= \frac{1}{\alpha}e^{\alpha v_0}<br /> so <br /> -\frac{1}{\alpha}\left(e^{\alpha v}- e^{\alpha v_0}\right)= -(A/m)t<br /> which is just what you have (except for the &quot;m&quot;).<br /> <br /> Are you concerned with solving for v?<br /> e^{\alpha v}- e^{\alpha v_0}= (A\alpha/m)t<br /> e^{\alpha v}= (A\alpha/m)t+ e^{\alpha v_0}<br /> Now take the logarithm of both sides.

 

[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 I am a little slow with this, am I making an error here?

 

[gabbagabbahey]

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

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...

 

[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.

 

[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

 

[gabbagabbahey]

Looks good to me

 

[littlehonda]

hey thanks a lot 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?