Kinematic Problem: How Long to Reach Half Speed in a Fluid?

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Homework Help Overview

The discussion revolves around a kinematic problem involving the motion of a marble in a fluid, where the acceleration is proportional to the speed of the marble. The original poster seeks assistance in determining the time required for the marble to reach half of its initial speed, given a specific acceleration equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of kinematic equations and the use of separation of variables in solving the problem. The original poster expresses difficulty in progressing beyond initial steps, while others suggest integrating the differential equation.

Discussion Status

There is an ongoing exchange of ideas, with some participants providing guidance on the mathematical approach. The original poster acknowledges confusion but shows engagement with the suggestions made. Multiple interpretations of the problem-solving process are being explored.

Contextual Notes

The problem involves specific conditions regarding the acceleration of the marble and its initial speed, which may influence the approach taken by participants. The original poster's attempts indicate a struggle with the mathematical formulation, particularly in applying calculus concepts.

bjon-07
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Hi all, can some one please help me on this problem;

The acceleration of a marble in a certain fluid is proportional to the speed of the marbles, and is given (in SI units) by a= -3.00V^2 for v>0. If it enters this fluid with a speed of 1.50 m/s, how long will it take for the marble to reach half its intial speed.


I have released that

Vf=Vi + a(t)

.75=1.5+ a(t)
-.15=a(t)


and that A=(Vf-VI)/t

I am stuck after this. Please help me.
 
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Do you calculus??This problem involves solving a first ODE through separation of variables...

Daniel.
 
I am still stock on this problem. I have tryed using separation of variables but it just leads me in a circle.
 
[tex]\frac{dv}{dt} = -3 v^2[/tex]
[tex]-\frac{1}{3 v^2} dv = dt[/tex]
Integrate!
 
I can't beilve I did not see that :eek: LOL, its soo easy. Thank you for your help.
 

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