Velocity Drag force and distance traveled

Click For Summary

Homework Help Overview

The discussion revolves around understanding the relationship between velocity, drag force, and distance traveled in the context of motion. Participants are exploring the application of differential equations to express force in terms of velocity and position.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to derive equations involving acceleration and velocity, questioning the dependency of variables and the proper substitutions needed. There is discussion about using the chain rule and expressing force as a differential equation.

Discussion Status

The conversation is active, with participants providing guidance on using the chain rule and expressing equations in terms of velocity and position. There is a focus on clarifying the definitions of variables and their dependencies, but no consensus has been reached on the specific formulations.

Contextual Notes

Participants are navigating the complexities of the problem, including the potential confusion around the definitions of initial and instantaneous velocity, as well as the relationships between the variables involved.

nysnacc
Messages
184
Reaction score
3

Homework Statement


Microsoft-Edge-Web-Notes-Mastering61302359[2305843009214067476].jpg


Homework Equations


Sum of force

a = dv/dt
v = integrate a + C

The Attempt at a Solution



I found the equation in terms of final v, initial v... however, on the right hand side, there is another v, and I think it is time dependent, so I am not sure which value should I plug in.
 
Physics news on Phys.org
Try using the chain rule to write ##a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt}##.
 
Then do we know dv/dx and dx/dt? on the right hand side, will the velocity be initial one? thanks
 
nysnacc said:
Then do we know dv/dx and dx/dt?
Try to express F = ma as a differential equation in terms of just the variables v and x. dv/dx is already expressed in these variables. How can you express dx/dt?

on the right hand side, will the velocity be initial one? thanks
v on the right side of a = F/m is the instantaneous velocity.
 
so dv/dx * dx/dt = -40/m v^2 ??
 
nysnacc said:
so dv/dx * dx/dt = -40/m v^2 ??
Yes. In order to get only the variables v and x to appear in the equation, what can you do with dx/dt?
 

Similar threads

Replies
57
Views
3K
Replies
10
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K