Acceleration in Fluid: Calculating Time for Marble's Speed to Halve

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The discussion centers on calculating the time it takes for a marble's speed to halve in a fluid where acceleration is proportional to the square of its speed, expressed as a = -3.00v^2. The initial speed of the marble is 1.50 m/s, and the challenge arises from the non-uniform acceleration, prompting the need for integration rather than standard kinematic equations. Participants suggest using the integral of acceleration to establish a velocity-time relationship, specifically integrating dv/v^2 = -3 dt. The integration leads to a formula that allows for solving the time when the speed is reduced to 0.75 m/s. The conversation highlights the necessity of understanding calculus concepts for solving this type of physics problem.
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Question states:

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

Here's what I think is it...take the derivative of a... a`=6v

d=6(.75m/s)
d=4.5m

I need the acceleration to sub in for the equation vf = vi + at
to find the time.

Am I doing any of this correct?
 
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Can't use that formula because this is not uniformily accelerated motion (acceleration is a function of velocity here). Try taking the integral of acceleration to get a v-t relationship.
 
remember from our last conversation that I don't know how to take integrals? :) Can you teach me with an example and explain what and integral does?
 
I could, but you should try reading about it on the internet, or better yet, in a calculus textbook, which will contain all the information you need. On a basic level, integration is a way to find areas under and between curves using only their equation. If you are not required to know calculus for your course, however, the questions are most likely doable without integration.

Sorry, I can't explain much more now. Gotta run.
 
UrbanX,

Since you didn't specify the mass of the marble I take it that you are ignoring gravity.

In that case
\frac {dv}{dt} = -3v^2
which amounts to integrating
\frac {dv}{v^2} = -3 dt
You should be able to handle it from there.
 
I actually have no clue what happened and how to even begin. Thank you for the example but I do not know what you are doing since I do not know how to integrate
 
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Taking from Tide's answer. Upon request by Urban.

\frac {dv}{v^2} = -3 dt

\int^{v}_{v_{o}} \frac {dv}{v^2} = \int^{t}_{0} -3 dt

-\frac{1}{v}]^{v}_{v_{o}} = -3t]^{t}_{0}


-\frac{1}{v} + \frac{1}{1.5}= -3t - 0

Urban, you know initial speed is 1.5, and v will be equal 1.5/2, so just solve for t.
 
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