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
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  • Explore the relationship between velocity, acceleration, and displacement
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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 :)
 

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