What is the Terminal Velocity of an Object Subjected to Drag Force?

AI Thread Summary
The discussion focuses on deriving the equations of motion for an object with a mass of 0.5 kg dropped from 500 meters, subjected to a drag force defined as fD = 2v. The terminal velocity is calculated to be 2.45 m/s, but participants struggle with explicitly deriving the functions v(t) and y(t) due to variable dependencies. A first-order linear differential equation approach is suggested to express the motion in terms of time. Participants clarify the importance of correctly defining variables and initial conditions, particularly ensuring that displacement y(t) starts from zero at t=0. The conversation highlights the need for careful mathematical manipulation and understanding of the physics involved.
Yapper
Messages
101
Reaction score
0

Homework Statement


An object of 0.5kg is dropped from rest at an altitude of h=500 meters. It
falls while being subjected to a drag force, defined as: fD = 2v .
Derive explicitly: v(t), and y(t) and plot v(t) given the initial conditions; use: t between 0
and 0.4 seconds, and v between 0 and -3 m/s
Based on your graph for v(t), what is the terminal velocity? Confirm this result
analytically (either using the limit or using Newton).

Homework Equations



a = dv/dt v = dy/dt

The Attempt at a Solution



For my equation i get 2v - mg = ma setting ma = 0 i solve for terminal velocity and get v = mg/2 = 2.45 m/s but I am having trouble explicitly deriving v(t) and y(t) because I end up with v = 9.8t - 4y which is 1 variable too many. How do I derive the equation for v with t as the only variable.
 
Physics news on Phys.org
Have you learned about differential equations?

Replacing a=dv/dt, write up the equation of motion in terms of v and t. It is a first-order linear de. Solve with the initial conditions given.

ehild
 
I've taken Calc I and II and I am taking Calc III right now and I never remember having to solve these. So far based on what I just been reading i get v(t) = 2.45 - 2.45e^4t with the initial condition being V(0) = 0. is that right? and if so why is the graph backwards. it seems like it should be 2.45 - 2.45e^-4t but I don't see where I made a mistake.

My work:

dv/dt - 4v = -9.8 I multiplied both sides by e^-4t to get v'e^-4t + -4e^-4t = -9.8e^-4t

so that gets me (ve^-4t)' = -9.8e^-4t

I integrate both sides then try to get v by itself and end up with:

v= 2.45 + ce^4t with v(0) = 0 c = -2.45
 
I changed my frame of reference around and it worked out better using dv/dt + 4v = 9.8.

I also got y(t) = 2.45t + 6.125e^-4t is that right?
 
You forgot to divide g by m. And clarify, what do you mean by y.
Edit: G has to be divided, not g.
ehild
 
Last edited:
why would I divide gravity by mass?

y is the displacement in the y direction from the point that the object began following
 
Well, you are right, I mixed it with weight...

If y(t) is displacement, it has to be zero at t=0.
Also, the original height is given. I think, the problem asks the height as function of t.

ehild
 
Back
Top