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Homework Help: Newtonian Mechanics

  1. Feb 11, 2004 #1
    A boat with initial speed v is launched on a lake. The boat is slowed by the water by a force, F=-ke^(bv). Find the expression for speed v(t).

    I've done the problem, but my answer seems too odd to be right...it may be my calculus.

    I've drawn a FBD, with the normal force and weight cancelling each other out. The net force is the resisting force, F=-ke^(bv), which I've then set equal to F=ma. I've used dv/dt for a.

    -ke^(bv)=m*(dv/dt)
    Rearranging to get like terms together gives me
    dv/(e^(bv))=-(k/m)dt
    (e^-(bv))dv=-(k/m)dt
    Setting up the integrand using limits 0 to t and initial v to v gives me:
    -b((e^-(bv.initial))-(e^-(bv)))=-(k/m)t
    Simplified to:
    (kt/bm)=e^(v.initial/v)
    ln (kt/bm)=(v.inital/v)
    So, v=(v.initial)/ln(kt/bm)

    The answer seems too messy...any help would be much appreciated!
     
  2. jcsd
  3. Feb 12, 2004 #2
    As to the calculus:
    How does
    -b((e^-(bv.initial))-(e^-(bv)))=-(k/m)t
    become
    (kt/bm)=e^(v.initial/v)
    ?
    You're saying that
    e^(-bv) - e^(-bvi) = e^(-vi/v)
    I don't think so.

    But, this is not a calculus problem anyway. I tried correcting your integration but still ended up with a nonsensical result until I realized that we shouldn't be forming a differential equation out of that force expression -- that's already the solution to a differential equation. Now, all you have to do is put it into an equation to describe the velocity, a la
    [tex]v(t) = v_0 + at[/tex]
    If
    [tex] f = ma[/tex]
    then
    [tex] a = \frac{f}{m} = -\frac{k}{m}e^{bv}[/tex]
    So,
    [tex]v(t) = v_0 - \frac{kt}{m}e^{bv}[/tex]
     
  4. Feb 12, 2004 #3

    HallsofIvy

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    gnome, that's hardly a solution- you still have ebv on the right side.

    mind circus, the only objection I have is that you have the integral of e-bv as -b-bv when it should be -(1/b)e-bv.

    Integrating e-bvdv= (-k/m)dt, I get -(1/b)e-bv= (-k/m)t+ C. Taking v= v0 when t=0, C= (-1/b)e-bv0 so
    e-bv= (bk/m)t+e-bv0 or
    v(t)= -(1/b)ln((bk/m)t+ e-bv0)
     
  5. Feb 15, 2004 #4
    Oops -- good point, Halls!

    But, I just went digging through my trash (which tends to accumulate), and this
    v(t)= -(1/b)ln((bk/m)t+ e-bv0)
    is the same as the result I got, which I had rejected because I thought it was nonsensical. As t gets big, this v takes on ever-increasing negative values. This defies reason. Logically, v should tend to 0.

    Actually, the expression for the force may be the problem. When v=0, F=-k, so the boat starts going backwards. F only approaches 0 asymptotically as v becomes a big enough negative number causing e^(bv) to approach 0.

    Or am I misunderstanding something?
     
  6. Feb 16, 2004 #5

    HallsofIvy

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    Science Advisor

    Good point. I strongly suspect that the force function was
    F=-ke(-bv). That is, that the exponential is negative, not positive.
     
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