Classical mechanics - finding distance D in terms of velocity

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Homework Help Overview

The problem involves a passenger stepping out of an airplane and falling under the influence of gravity and air resistance, modeled as a linear function of velocity. The goal is to find the distance fallen in terms of velocity.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the "vdv/dx trick" to eliminate time from the equations of motion. There are attempts to simplify the equations by eliminating mass and isolating variables. Questions arise regarding the integration process and the appropriate limits for integration.

Discussion Status

Participants are actively engaging with the problem, providing suggestions for simplification and variable separation. There is a focus on the integration step, with some guidance offered regarding the limits based on the problem statement.

Contextual Notes

The problem specifies that the passenger starts from rest and provides a clear setup for the forces involved, which participants are referencing in their attempts to solve the problem.

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Homework Statement



"A passenger (mass m) initially at rest steps out of an airplane. Assume down is the positive x-axis and put the origin at the airplane. Assume the air resistance force is linear in the velocity so F(air)= -mbv. Find the distance D he has fallen when his velocity is v."

Homework Equations



Equations of motion

The "vdv/dx trick": d2z/dt2 = (z-dot)*(d(z-dot)/dz)

F(tot) = ma

F(air) = -mbv

Weight = mg

The Attempt at a Solution



Here's how far I've gotten:

Since the skydiver is only falling in the x direction, there's only one equation of motion, which I found to be ma = -mbv + mg [or, alternatively, m(x-double dot) = -mb(x-dot) + mg]. Now, I know I want the relation of distance and velocity, without time, so I use the "vdv/dx" trick (so that there's no longer time in the equation).

That makes this mv*(dv/dx) = -mbv + mg, or m(x-dot)*(d(x-dot)/dx) = -mb(x-dot) + mg. I rearranged this to get dx = (-m/b)*(vdx/v-(mg/b)), where -(mg/b) is the terminal velocity.
I'm sorry for all the writing, but am I correct so far? And how do I continue to solve the problem from here? Any help would be appreciated.
 
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First off, divide through by m to simplify the expression a bit. Secondly, can you separate variables, i.e have things with "v" on one side and "x" on the other?
 
First, since m is common to every term, I would eliminate that to make things easier. Next, you have vdv/dx = g-bv. Try to isolate v and dv on one side, and dx on the other side.
 
Alright, so...

I went back and simplified what I had first and ended up with v(dv/dx) = -bv - g. So, after isolating dz, I end up with dz = -(vdv/(bv+g)).

Now, I know my next step is to integrate this, but I'm not sure what the limits would be.
 
The problem tells you what the limits should be.

"A passenger (mass m) initially at rest steps out of an airplane. Assume down is the positive x-axis and put the origin at the airplane. Assume the air resistance force is linear in the velocity so F(air)= -mbv. Find the distance D he has fallen when his velocity is v."
 

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