Solving x(t) using F(v) for Particle of Mass m

  • Thread starter Thread starter Tonyt88
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around finding the position function x(t) for a particle of mass m subjected to a force defined by F(v) = -m(alpha)v^2. The initial conditions include a starting position of zero and an initial speed denoted as v nought.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss solving the differential equation derived from the force equation, with some expressing uncertainty about their steps and the integration process. Questions arise regarding the handling of constants of integration and the correctness of derived expressions for velocity and position.

Discussion Status

The discussion includes attempts to derive expressions for velocity and position, with some participants providing feedback on each other's reasoning. While there is no explicit consensus, guidance has been offered regarding the integration process and the need for constants in the solutions.

Contextual Notes

Some participants indicate confusion about the steps in solving the differential equations and the implications of initial conditions. There is also mention of potential missing information regarding integration constants.

Tonyt88
Messages
62
Reaction score
0
A particle of mass m is subject to a force F(v) = -m(alpha)v^2. The initial position is zero, and the initial speed is v nought. Find x(t).
 
Physics news on Phys.org
It's only a matter of solving the differential equation

[tex]-m\alpha v^2=m\frac{dv}{dt}[/tex]

and then

[tex]v=\frac{dx}{dt}[/tex]
 
Sorry perhaps I'm not catching on, so I have:

-1/v = -(alpha)t

Do I merely say that 1/(alpha t) = dx/dt or perhaps I'm missing something.
Or is my first step incorrect?
 
You forgot the constant of integration. The solution to the differential equation for v is

-(alpha)t = -1/v +C

And plugging v(0)=[itex]v_0[/itex] gives C=1/[itex]v_0[/itex]. So

[tex]v(t)=\frac{1}{\alpha t}+\frac{1}{v_0}[/tex]

And now your have to solve the differential equation.

[tex]\frac{dx}{dt}= \frac{1}{\alpha t}+\frac{1}{v_0}[/tex]

with initial condition x(0)=0 to find x(t).

Makes sense?
 
Okay, so I got:

x(t) = (ln(t)/alpha) + (t/v nought) + C
and C = 0 at x(0)

So I have x(t) = (ln(t)/alpha) + (t/v nought)

Is that correct or did I once again miss something?
 
It looks fine.
 

Similar threads

Replies
3
Views
3K
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
8
Views
3K