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

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SUMMARY

The discussion centers on deriving the acceleration of a particle in motion described by the equation s = kv²ln(v), where k is a constant and v represents velocity. The solution provided is a = 1/(k(1 + 2ln(v))). Participants emphasize the application of the Chain Rule and the relationship between acceleration, velocity, and position. Key methods include differentiating with respect to time and utilizing the relationship a = dv/dt = (dv/dx)(dx/dt) = v(dv/dx) to find acceleration in terms of velocity.

PREREQUISITES
  • Understanding of calculus, specifically differentiation and the Chain Rule.
  • Familiarity with the concepts of velocity and acceleration in physics.
  • Knowledge of logarithmic functions and their properties.
  • Ability to manipulate equations involving constants and variables.
NEXT STEPS
  • Study the application of the Chain Rule in calculus.
  • Learn about the relationship between velocity and acceleration in kinematics.
  • Explore logarithmic differentiation techniques for complex functions.
  • Investigate the implications of constants in motion equations, particularly in physics.
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the mathematical foundations of motion, particularly in deriving relationships between velocity and acceleration.

Warr
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Here's the question

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

s = kv^2ln v

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 a = \frac {1}{k(1+2ln v)}

Thanks in advance, Warr
 
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Apply the Chain Rule

Let u = v^2\ln{v} and then use the chain rule (and remember that v = ds/dt and a = dv/dt).
 
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...?
 
Relativistic Momentum, Mass, and Energy Momentum and mass (...), the classic equations for conserving momentum and energy are not adequate for the analysis of high-speed collisions. (...) The momentum of a particle moving with velocity ##v## is given by $$p=\cfrac{mv}{\sqrt{1-(v^2/c^2)}}\qquad{R-10}$$ ENERGY In relativistic mechanics, as in classic mechanics, the net force on a particle is equal to the time rate of change of the momentum of the particle. Considering one-dimensional...

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