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

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Homework Help Overview

The discussion revolves around a Newtonian mechanics problem involving a boat's speed as it is affected by a resistive force proportional to an exponential function of its speed. The original poster attempts to derive an expression for speed v(t) based on the given force equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various integration steps and the validity of the derived expressions. Questions are raised about the correctness of the integration process and the interpretation of the force function. Some participants suggest that the original force expression may be misinterpreted.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's reasoning and calculations. There is a recognition of potential issues with the force expression, and some participants express doubt about the results derived from the current formulation.

Contextual Notes

Participants are exploring the implications of the force function and its impact on the velocity expression. There is a suggestion that the exponential term in the force equation may need to be reconsidered, indicating a possible misunderstanding of the problem setup.

mindcircus
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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
[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]
 
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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