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!

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.

    Rearranging to get like terms together gives me
    Setting up the integrand using limits 0 to t and initial v to v gives me:
    Simplified to:
    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
    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]
    [tex] f = ma[/tex]
    [tex] a = \frac{f}{m} = -\frac{k}{m}e^{bv}[/tex]
    [tex]v(t) = v_0 - \frac{kt}{m}e^{bv}[/tex]
  4. Feb 12, 2004 #3


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    Good point. I strongly suspect that the force function was
    F=-ke(-bv). That is, that the exponential is negative, not positive.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook