How Long Does It Take for Speed to Halve with Non-Constant Acceleration?

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Homework Help Overview

The problem involves an object in a fluid where the acceleration is proportional to the square of its speed. The original poster seeks to determine the time it takes for the object's speed to reduce from 1.13 m/s to 0.565 m/s.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between acceleration and velocity, with attempts to integrate the acceleration formula. Questions arise regarding the interpretation of variables and the relevance of displacement versus time in the context of the equations used.

Discussion Status

Some participants have provided guidance on separating variables and integrating, while others are exploring the implications of using specific equations of motion. There is an ongoing examination of the assumptions related to acceleration and its derivation from velocity.

Contextual Notes

There appears to be confusion regarding the variables in the equations, particularly whether "x" refers to time or distance, which may affect the interpretation of the problem. The discussion reflects the complexity of non-constant acceleration scenarios.

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Homework Statement



The acceleration of an object in a fluid is proportional to its speed squared. If the object's initial speed is 1.13 m/s, how much time until its speed is reduced by half (0.565 m/s)?

Homework Equations



a=-.48v^2

The Attempt at a Solution



I tried using the formula

v(f)-v(0) + a*t

Taking the derivative of the whole thing seems like it's fruitless. If I integrate the formula I can get an equation that relates to velocity:

v=-.16x^3

But I don't know how that relates to the time. Should the "x" be displacement or time since velocity (the result of the equation) is displacement AND time? I can solve for x, but I don't know what I'm actually solving. thanks everyone
 
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Your equation relating acceleration to velocity can be written as

[tex]a=\frac{dv}{dt}=kv^2[/tex]

Separate the variables and integrate.
 
miggitymark said:
I tried using the formula

v(f)-v(0) + a*t

Remember that these quick little equations of motion most of us memorize are all derived from the case of constant acceleration!
 
Ahh, the acceleration IS the derivative. So the integral with -.16 is the equation I need to use? What is the X in that equation then? time or distance?
 
Separating variables gives

[tex]\frac{dv}{v^2}=kdt[/tex]

Integrate both sides using the appropriate limits for v (v0 to v0/2) and t (0 to t).
 

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