Ordinary differential equations with boat

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Homework Help Overview

The discussion revolves around a problem involving ordinary differential equations related to a boat's motion after power is shut off. The scenario includes water resistance that is proportional to the velocity raised to a constant power, prompting participants to explore the relationship between velocity and distance traveled.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the setup of the force balance equation and the implications of water resistance on the boat's motion. There are inquiries about deriving velocity as a function of distance rather than time, leading to discussions on integration and the relationship between velocity and distance.

Discussion Status

Some participants have provided insights into the mathematical formulation of the problem, while others have raised questions about the necessity of expressing velocity in terms of distance. There is an ongoing exploration of how to transition from velocity as a function of time to distance.

Contextual Notes

Participants are navigating the complexities of the problem setup, including the assumptions about resistance and the initial conditions provided. There is a noted confusion regarding the requirement to express velocity as a function of distance, which is central to the problem.

Anabelle37
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Question:
A boat of mass m is traveling with the velocity v0. At t=0 the power is shut off. Assuming water resistance proportional to v^n, where n is a constant and v is the instantaneous velocity, find v as a function of the distance travelled. (Note that you need to consider the two cases).

I'm having trouble setting the problem up. Do I need to have a force balance equation? What would it be as the one I originally thought of doesn't make much sense: mdv/dt=v0 - kv^n.

Please help asap!
Thanks
 
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After the power is shut off, the only force on the boat is the resistance, so your equation is m*dv/dt = -kvn. v0 is the velocity at time t = 0: it's the initial condition.
 
Ok, thanks heaps.
How do I solve for v(x) not v(t)??

I've started:
dv/dt= -(k/m)v^n
dv/dt= -av^n where a =k/m
integral (1/v^n) dv = intgeral (-a) dt
(v^(-n+1))/-n+1= -at + c
?
 
Or you could just attach a lightweight floating plumb line, such as polypropylene, paying it out for about three minutes until it comes to rest, and backwards solve...

Sorry for the intrustion, folks. I've been out of academia and in the real world way too long. I'll let you be.

I would ask you, though, if you'd like some ideas on how to model situations so as to obtain data required for parameterized equations required for backsolving solutions, I'd be happy to work with you. I'm sort of interested in getting back into anything having to do with Dffy-Q's - it was the only higher math class in college I in which I earned a solid A.

As for the physical situations upon which they're based, I've experienced them in droves, and on many different scales, from micro-robotics to aerial transport.
 
Anabelle37 said:
Ok, thanks heaps.
How do I solve for v(x) not v(t)??
Why do you think you want v(x)? Velocity is usually in terms of time, not distance.
Anabelle37 said:
I've started:
dv/dt= -(k/m)v^n
dv/dt= -av^n where a =k/m
integral (1/v^n) dv = intgeral (-a) dt
(v^(-n+1))/-n+1= -at + c
?
So v^(-n + 1) = (n - 1)(at - c'), where c' = -c

To solve for v, raise each side to the power 1/(1 - n).
 
Ok thanks heaps.
"Why do you think you want v(x)? Velocity is usually in terms of time, not distance."[/I-mark
Because the question says find v as a function of the distance travelled?
 
Sorry, I overlooked that piece of information.

I think this would work - You have v as a function of t. If you integrate v (with respect to t) you get distance as a function of t. From v = v(t) and s = s(t), you should be able to solve algebraically for v as a function of distance.
 

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