# Newtonian Mechanics

1. Feb 11, 2004

### mindcircus

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. Feb 12, 2004

### gnome

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
$$v(t) = v_0 + at$$
If
$$f = ma$$
then
$$a = \frac{f}{m} = -\frac{k}{m}e^{bv}$$
So,
$$v(t) = v_0 - \frac{kt}{m}e^{bv}$$

3. Feb 12, 2004

### HallsofIvy

Staff Emeritus
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)

4. Feb 15, 2004

### gnome

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?

5. Feb 16, 2004

### HallsofIvy

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