1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    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

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

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

    And

    [tex]\int-Adt=-At+C_2[/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

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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")
    Integrating,
    [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]
    so
    [tex]C= \frac{1}{\alpha}e^{\alpha v_0}[/tex]
    so
    [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: Feb 6, 2009
  9. Feb 6, 2009 #8
    [itex]v_0=v(t=0)[/itex]

    (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

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    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

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook