What is the Simplified Equation for Resistive Force Acting on a Speed Skater?

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Homework Help Overview

The discussion revolves around the resistive force acting on a speed skater, modeled as proportional to the square of the skater's speed. The original poster presents a specific equation for the skater's speed over time after crossing the finish line, seeking assistance in deriving and understanding the equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the formulation of a separable differential equation based on the resistive force. Questions arise regarding the introduction of initial conditions and the integration process, with some participants discussing the necessary steps to arrive at the final equation.

Discussion Status

The discussion is active, with participants providing insights into the mathematical formulation and integration steps. Some guidance has been offered regarding the handling of initial conditions, and there is an acknowledgment of potential errors in the simplification process.

Contextual Notes

Participants are working within the constraints of a homework assignment, focusing on the derivation of equations without providing complete solutions. There is an emphasis on understanding the mathematical relationships involved.

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Consider an object which the net force is a resistive force proportional to the square of its speed. For example: assume that the resistive force acting on a speed skater is F=-k*m*V^2, where k is a constant and m is the skater's mass. The skater crosses the finish line of a straight-line race with speed V(i) and the slows down by coasting on his skates. Show that his speed at time "t", any time after the finish line is equal to Vf=Vi/(1+Vi*k*t).

Any suggestions? Thank you in advance!
 
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It's a separable differential equation.
Go ahead!
 
You mean mdv/dt= -mkV^2 cancel m, gives dv/dt= -kv^2, which in turn yields dv=-kv^2dt? But how does the Vi gets introduced?

Thank you
 
You get:
[tex]\frac{dv}{v^{2}}=-kdt[/tex]
Right?
Vi enters as the initial condition in that v(0)=Vi
 
Thank you. I know I have to integrate both sides. Left side from Vi to V(t) and the right side from t to 0. I get -kt for the right side. But I am having trouble with the left.
 
Post what you've gotten so far! (The equation)
 
I got it. Thank you. 1/V(t)-1/Vi=-kt. Than simplify...
 
You have a sign flaw; you should simplify:
[tex]-\frac{1}{V(t)}+\frac{1}{V_{i}}=-kt[/tex]
 

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