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Finding v(t) for a velocity dependent force

  1. Feb 6, 2009 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    F = m(dV/dt)
    A and [tex]\alpha[/tex] 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^(-[tex]\alpha[/tex]v)t

    or should the v be eliminated from the right hand side. I tried doing that by dividing the left by e^(-[tex]\alpha[/tex]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. jcsd
  3. Feb 6, 2009 #2


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    Hint: You have a separable differential equation for v(t):wink:
  4. Feb 6, 2009 #3
    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^([tex]\alpha[/tex]v) dv = -Adt

    1/[tex]\alpha[/tex](e^([tex]\alpha[/tex]v)-e^([tex]\alpha[/tex]v0)) = -At

    i'm not sure what this means though or where to go from here
  5. Feb 6, 2009 #4


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    [tex]\int e^{\alpha v}dv =\frac{e^{\alpha v}}{\alpha}+C_1[/tex]



    [tex]\implies \frac{e^{\alpha v}}{\alpha}=-At+C_3[/tex] where [itex]C_3 \equiv C_1-C_2[/itex]

    Multiply both sides of the equation by [itex]\alpha[/itex] and then take the natural log. What initial conditions are you given?
  6. Feb 6, 2009 #5
    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
    [tex]\alpha[/tex]=.1 s/m
    v0= 20 m/s
  7. Feb 6, 2009 #6


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    I assume [itex]v_0=v(t=0)[/itex]? If so, you can solve for [itex]C_3[/itex].
  8. Feb 6, 2009 #7


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

    Are you concerned with solving for v?
    [tex]e^{\alpha v}- e^{\alpha v_0}= (A\alpha/m)t[/tex]
    [tex]e^{\alpha v}= (A\alpha/m)t+ e^{\alpha v_0}[/tex]
    Now take the logarithm of both sides.
    Last edited by a moderator: Feb 6, 2009
  9. Feb 6, 2009 #8

    (e^20[itex]\alpha[/itex])/[itex]\alpha[/itex] = C3

    C3 = 73.8

    v(t) = (ln(-At[itex]\alpha[/itex])+ ln(73.8[itex]\alpha[/itex]))/[tex]alpha[/tex]

    sorry if im a little slow with this, am I making an error here?
  10. Feb 6, 2009 #9


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    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

    [tex]\frac{me^{\alpha v}}{\alpha}=-At+C_3[/tex]

    Multiply both sides of the equation by alpha/m:

    [tex]\implies e^{\alpha v}=\frac{\alpha}{m}(-At+C_3)[/tex]

    Then take the natural log of both sides

    [tex]\alpha v =\ln\left(\frac{\alpha}{m}(-At+C_3)\right)[/tex]

    You can take it from there...
  11. Feb 6, 2009 #10
    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.
  12. Feb 6, 2009 #11
    v= (ln([itex]\alpha[/itex](-At+C3)))/[itex]\alpha[/itex]

    Solving for C3

    C3 = (e^v0[itex]\alpha[/itex])/[itex]\alpha[/itex]

    at V(0) t=0

    C3 = 73.89

    v= (ln([itex]\alpha[/itex](-At+73.89)))/[itex]\alpha[/itex]
  13. Feb 6, 2009 #12


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    Looks good to me :approve:
  14. Feb 6, 2009 #13
    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?
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