Solving Newtonian Mechanics: Find Expression for Speed v(t)

AI Thread Summary
The discussion revolves around deriving the expression for the speed v(t) of a boat experiencing a resistive force due to water, given by F=-ke^(bv). Participants analyze their calculations and integration steps, expressing concerns about the validity of their results, which seem counterintuitive. The consensus suggests that the force expression might be incorrectly formulated, as it leads to nonsensical outcomes where speed becomes increasingly negative. A proposed correction is to consider the force as F=-ke^(-bv) instead, which aligns better with expected physical behavior. Ultimately, the conversation highlights the importance of accurately defining the force function in solving the differential equation for speed.
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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!
 
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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}
 
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)
 
gnome, that's hardly a solution- you still have e^(bv) on the right side.
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?
 
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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