Equation of Motion for a Particle: Finding Acceleration in Terms of Velocity

  • Context: Undergrad 
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Discussion Overview

The discussion revolves around finding an expression for acceleration in terms of velocity for a particle described by the equation of motion s = kv^2ln v, where k is a constant. The scope includes mathematical reasoning and application of calculus concepts such as differentiation and the chain rule.

Discussion Character

  • Mathematical reasoning, Technical explanation, Homework-related

Main Points Raised

  • One participant presents the problem and requests assistance in deriving the acceleration from the given equation of motion.
  • Another participant suggests applying the chain rule and defines u = v^2ln v, indicating the relationship between velocity and acceleration.
  • A different participant advises differentiating both sides of the equation with respect to time and notes that this will relate velocity and acceleration.
  • One participant acknowledges the reminder that dv/dt represents acceleration.
  • Another participant proposes an alternative method involving the relationship a = dv/dt = (dv/dx)(dx/dt) = v(dv/dx) and suggests finding dx/dv to derive the necessary expression.
  • A later reply raises a question regarding the scenario when v = 0, indicating a potential point of discussion or concern.

Areas of Agreement / Disagreement

Participants present various methods to approach the problem, and while there is a general agreement on the need to differentiate and apply the chain rule, no consensus is reached on the final expression for acceleration or the implications of v = 0.

Contextual Notes

The discussion does not resolve the implications of the velocity being zero, and there may be assumptions regarding the behavior of the function at that point that are not fully explored.

Warr
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Here's the question

The equation of motion of a particle moving in a straight line is:

[tex]s = kv^2ln v[/tex]

where k is a constant and v is the velocity. Find an equation that expresses the acceleration in terms of velocity.

I need some help on this problem. I'd post my work but I don't exactly have time, and I need to know how to do this by tomorrow morning.

The answer is [tex]a = \frac {1}{k(1+2ln v)}[/tex]

Thanks in advance, Warr
 
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Apply the Chain Rule

Let [itex]u = v^2\ln{v}[/itex] and then use the chain rule (and remember that [itex]v = ds/dt[/itex] and [itex]a = dv/dt[/itex]).
 
Differentiate both sides of the equation with respect to t.

Note that the left hand side will turn into velocity, and the right hand side will turn into some function of v and dv/dt (i.e. acceleration).

Now solve for a.

cookiemonster
 
Thanks guys, I completely forgot that dv/dt was a!
 
There's another neat way to do this.

Notice that a=dv/dt=(dv/dx)(dx/dt)=v(dv/dx)

Since you have x=f(v), find dx/dv and invert it to get dv/dx. Multiply this by v and you have your answer !
 
So when v = 0...?
 

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