Air friction modeled as a differential equation

[DoubleWs]

1. Find 'y' in terms of 't' for the equation of the net force on a falling object ƩFy = mg-bv = ma
So I'm pretty sure I found the velocity in terms of time correctly, and I'm interested in having someone check my answer for 'y' in terms of time.

Homework Equations

ƩFy = mg-bv = ma
a = $\frac{dv}{dt}$
b = a constant (drag coefficient)

The Attempt at a Solution

mg - bv = ma

mg - bv = m$\frac{dv}{dt}$

g - $\frac{bv}{m}$ = $\frac{dv}{dt}$

(g - $\frac{bv}{m}$)dt = dv

∫dt = ∫$\frac{dv}{g-(bv/m)}$

g - $\frac{bv}{m}$ = u

$\frac{du}{dv}$ = $\frac{-b}{m}$

$\frac{-m}{b}$du = dv

∫dt = -$\frac{m}{b}$∫$\frac{1}{u}$du

t + C = $\frac{-m}{b}$ln(u)

t + C = $\frac{-m}{b}$ln(g - $\frac{bv}{m}$)

$\frac{-bt}{m}$ + C = ln(g - $\frac{bv}{m}$)

e$^{-bt/m}$ * C = g - $\frac{bv}{m}$

$\frac{mg}{b}$ - C * e$^{-bt/m}$ = v

$\frac{mg}{b}$(1 - e$^{-bt/m}$) = v

Ok, so there's 'v' in terms of 't'. Now...

$\frac{mg}{b}$(1 - e$^{-bt/m}$) = $\frac{dy}{dt}$

∫$\frac{mg}{b}$(1 - e$^{-bt/m}$)dt = ∫dy

$\frac{mg}{b}$∫(1 - e$^{-bt/m}$)dt = ∫dy

$\frac{mg}{b}$(t + $\frac{m}{b}$e$^{-bt/m}$) - C = y

$\frac{mg}{b}$(t + $\frac{m}{b}$e$^{-bt/m}$) - 1 = y

Just wondering if all of this is right, and if now, how can I fix it.

 

[collinsmark]

Hello DoubleWs,

Welcome to Physics Forums!

1. Find 'y' in terms of 't' for the equation of the net force on a falling object ƩFy = mg-bv = ma
So I'm pretty sure I found the velocity in terms of time correctly, and I'm interested in having someone check my answer for 'y' in terms of time.

Homework Equations

ƩFy = mg-bv = ma
a = $\frac{dv}{dt}$
b = a constant (drag coefficient)

The Attempt at a Solution

mg - bv = ma

mg - bv = m$\frac{dv}{dt}$

g - $\frac{bv}{m}$ = $\frac{dv}{dt}$

(g - $\frac{bv}{m}$)dt = dv

∫dt = ∫$\frac{dv}{g-(bv/m)}$

g - $\frac{bv}{m}$ = u

$\frac{du}{dv}$ = $\frac{-b}{m}$

$\frac{-m}{b}$du = dv

∫dt = -$\frac{m}{b}$∫$\frac{1}{u}$du

t + C = $\frac{-m}{b}$ln(u)

t + C = $\frac{-m}{b}$ln(g - $\frac{bv}{m}$)

$\frac{-bt}{m}$ + C = ln(g - $\frac{bv}{m}$)

e$^{-bt/m}$ * C = g - $\frac{bv}{m}$

$\frac{mg}{b}$ - C * e$^{-bt/m}$ = v

Everything up to this point looks pretty good. At least it's the same result that I came up with.

$\frac{mg}{b}$(1 - e$^{-bt/m}$) = v

But here you're making a big assumption. The above relationship is only true if there is a boundary condition (initial condition) that states v = 0 at time t = 0. There is no such boundary condition in the problem statement.

It might not be a bad assumption though, all I'm saying is the problem statement never mentions it. (There probably should be some sort of boundary conditions though, because if the initial velocity was in the opposite direction [heading up], none of the above would apply for that situation, because the differential equation would become ƩFy = mg+bv = ma until the object turned around.)

All your work below is contingent upon this assumption, so I'll treat everything below as though v = 0 at time t = 0.

Ok, so there's 'v' in terms of 't'. Now...

$\frac{mg}{b}$(1 - e$^{-bt/m}$) = $\frac{dy}{dt}$

∫$\frac{mg}{b}$(1 - e$^{-bt/m}$)dt = ∫dy

$\frac{mg}{b}$∫(1 - e$^{-bt/m}$)dt = ∫dy

$\frac{mg}{b}$(t + $\frac{m}{b}$e$^{-bt/m}$) - C = y

Everything seems to check out so far.

$\frac{mg}{b}$(t + $\frac{m}{b}$e$^{-bt/m}$) - 1 = y

Something is a bit wrong with the above. I'm assuming here that you have assumed that y = 0 at time t = 0. But even with that assumption, something is still not right. It could be simple matter of missing parenthesis and/or parenthesis not in the right place. But as the equation stands now, it's not dimensionally correct (the y has units of length, and the 1 has no units at all); and y ≠ 0 at time t = 0, which is what I'm guessing you wanted to assume in the first place.