Particle Velocity and Distance in a Straight Line with -2v Acceleration

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SUMMARY

The discussion focuses on a particle moving in a straight line with an acceleration defined by the equation a = -2v m/s². Given initial conditions of v = 20 m/s at s = 0 and t = 0, the objective is to derive the particle's velocity as a function of position and determine the total distance traveled before coming to a stop. The solution involves using the relationship a = v (dv/ds) to express velocity in terms of displacement, leading to the integration of the velocity function.

PREREQUISITES
  • Understanding of kinematics equations
  • Familiarity with calculus, specifically integration techniques
  • Knowledge of differential equations
  • Concept of velocity as a function of displacement
NEXT STEPS
  • Study the derivation of velocity functions from acceleration equations
  • Learn about integrating differential equations in physics
  • Explore the relationship between velocity, acceleration, and displacement
  • Investigate real-world applications of kinematic equations in motion analysis
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This discussion is beneficial for physics students, educators, and anyone interested in understanding motion dynamics and kinematic equations in a straight line context.

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Homework Statement


A particle is moving along a straight line such that its acceleration is defined as a = (-2v) m/s^2, where v is in meters per second. If v = 20 m/s when s = 0 and t = 0, determine the particle's velocity as a function of position, and the distance the particle moves before it stops.


Homework Equations


Basic kinematics equations


The Attempt at a Solution


I tried using dt = dv/a and integrating that, and I got t = (ln20 - lnv)/2, then I tried substituting it into s = s0 + v0t + (1/2)at^2, and now I'm stuck. Any help would be appreciated. Thanks :)
 
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Use the fact that

a= \frac{dv}{dt}=v \frac{dv}{ds}

this will help you get velocity in terms of displacement.
 
Oh right. Thanks a bunch :)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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