- #1
DoubleWs
- 1
- 0
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.
ƩFy = mg-bv = ma
a = [itex]\frac{dv}{dt}[/itex]
b = a constant (drag coefficient)
mg - bv = ma
mg - bv = m[itex]\frac{dv}{dt}[/itex]
g - [itex]\frac{bv}{m}[/itex] = [itex]\frac{dv}{dt}[/itex]
(g - [itex]\frac{bv}{m}[/itex])dt = dv
∫dt = ∫[itex]\frac{dv}{g-(bv/m)}[/itex]
g - [itex]\frac{bv}{m}[/itex] = u
[itex]\frac{du}{dv}[/itex] = [itex]\frac{-b}{m}[/itex]
[itex]\frac{-m}{b}[/itex]du = dv
∫dt = -[itex]\frac{m}{b}[/itex]∫[itex]\frac{1}{u}[/itex]du
t + C = [itex]\frac{-m}{b}[/itex]ln(u)
t + C = [itex]\frac{-m}{b}[/itex]ln(g - [itex]\frac{bv}{m}[/itex])
[itex]\frac{-bt}{m}[/itex] + C = ln(g - [itex]\frac{bv}{m}[/itex])
e[itex]^{-bt/m}[/itex] * C = g - [itex]\frac{bv}{m}[/itex]
[itex]\frac{mg}{b}[/itex] - C * e[itex]^{-bt/m}[/itex] = v
[itex]\frac{mg}{b}[/itex](1 - e[itex]^{-bt/m}[/itex]) = v
Ok, so there's 'v' in terms of 't'. Now...
[itex]\frac{mg}{b}[/itex](1 - e[itex]^{-bt/m}[/itex]) = [itex]\frac{dy}{dt}[/itex]
∫[itex]\frac{mg}{b}[/itex](1 - e[itex]^{-bt/m}[/itex])dt = ∫dy
[itex]\frac{mg}{b}[/itex]∫(1 - e[itex]^{-bt/m}[/itex])dt = ∫dy
[itex]\frac{mg}{b}[/itex](t + [itex]\frac{m}{b}[/itex]e[itex]^{-bt/m}[/itex]) - C = y
[itex]\frac{mg}{b}[/itex](t + [itex]\frac{m}{b}[/itex]e[itex]^{-bt/m}[/itex]) - 1 = y
Just wondering if all of this is right, and if now, how can I fix it.
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 = [itex]\frac{dv}{dt}[/itex]
b = a constant (drag coefficient)
The Attempt at a Solution
mg - bv = ma
mg - bv = m[itex]\frac{dv}{dt}[/itex]
g - [itex]\frac{bv}{m}[/itex] = [itex]\frac{dv}{dt}[/itex]
(g - [itex]\frac{bv}{m}[/itex])dt = dv
∫dt = ∫[itex]\frac{dv}{g-(bv/m)}[/itex]
g - [itex]\frac{bv}{m}[/itex] = u
[itex]\frac{du}{dv}[/itex] = [itex]\frac{-b}{m}[/itex]
[itex]\frac{-m}{b}[/itex]du = dv
∫dt = -[itex]\frac{m}{b}[/itex]∫[itex]\frac{1}{u}[/itex]du
t + C = [itex]\frac{-m}{b}[/itex]ln(u)
t + C = [itex]\frac{-m}{b}[/itex]ln(g - [itex]\frac{bv}{m}[/itex])
[itex]\frac{-bt}{m}[/itex] + C = ln(g - [itex]\frac{bv}{m}[/itex])
e[itex]^{-bt/m}[/itex] * C = g - [itex]\frac{bv}{m}[/itex]
[itex]\frac{mg}{b}[/itex] - C * e[itex]^{-bt/m}[/itex] = v
[itex]\frac{mg}{b}[/itex](1 - e[itex]^{-bt/m}[/itex]) = v
Ok, so there's 'v' in terms of 't'. Now...
[itex]\frac{mg}{b}[/itex](1 - e[itex]^{-bt/m}[/itex]) = [itex]\frac{dy}{dt}[/itex]
∫[itex]\frac{mg}{b}[/itex](1 - e[itex]^{-bt/m}[/itex])dt = ∫dy
[itex]\frac{mg}{b}[/itex]∫(1 - e[itex]^{-bt/m}[/itex])dt = ∫dy
[itex]\frac{mg}{b}[/itex](t + [itex]\frac{m}{b}[/itex]e[itex]^{-bt/m}[/itex]) - C = y
[itex]\frac{mg}{b}[/itex](t + [itex]\frac{m}{b}[/itex]e[itex]^{-bt/m}[/itex]) - 1 = y
Just wondering if all of this is right, and if now, how can I fix it.
Last edited: