Solve Maximum Velocity: Determine a, x, k, v for Particle

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Homework Help Overview

The problem involves a particle experiencing a position-dependent acceleration described by the equation a = k/(x+4)². The particle starts from rest at the origin, and the task is to determine the constant k, the position at which the velocity reaches 4.5 m/s, and the maximum velocity of the particle when it reaches a velocity of 4 m/s at x = 8 m.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between acceleration and velocity, particularly questioning how to determine the maximum velocity based on the acceleration equation. There is an exploration of integrating the acceleration to find velocity as a function of position and subsequently time.

Discussion Status

Some participants have made progress on parts of the problem, particularly in calculating values for k and the position at a specific velocity. However, there is uncertainty regarding the approach to finding the maximum velocity, with discussions about the implications of setting acceleration to zero and the nature of the integrals involved.

Contextual Notes

Participants are navigating the complexities of integration and the interpretation of results, with some expressing difficulty in handling the resulting integrals and their implications for maximum velocity.

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Homework Statement



A particle starts from rest at the origin and is given an acceleration a=k/(x+4)^2, where a and x are expressed in m/s2 and m, respectively, and k is a constant. Knowing that the velocity of the particle is 4 m/s when x=8m, determine (a) the value of k, (b) the position of the particle when v = 4.5 m/s, (c) the maximum velocity of the particle.

Homework Equations





The Attempt at a Solution



I calculated part (a) and (b) easily. However, I am having some troubles with part (c).
The maximum velocity occurs when the acceleration is equal to zero because there is either a minimum or maximum right? So I set

0 = k/(x+4)^2

which cannot be.
What am I doing wrong?
 
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write:

[tex] v\frac{dv}{dx}=\frac{k}{(x+4)^{2}}[/tex]

Integrate to get v as a function of x and then write v=dx/dt so find x as a function of t then differentiate again and examine the limit as t tends to infinity.
 
I am getting a very ugly integral. But you said to get v as function of x right?
Then

v = dx/dt ---> dt = (1/v)dx

and integrate the above to get the position x as a function of time? And then take the limit of that function as t goes to infinity? Why would that give me the maximum velocity? Is it because the graph of x(t) has a vertical asymptote?
 
What do you get for the integral?
 

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