- #1
dustbin
- 240
- 5
Homework Statement
I need to integrate
[tex] v(t) = V( \frac{1- e^{-2gt/V}}{1+ e^{-2gt/V}}) [/tex]
to show that the position function is given by
[tex] s(t) = Vt + \frac{V^2}{g}ln(\frac{1 + e^{-2gt/V}}{2}) [/tex]
Homework Equations
g is the acceleration due to gravity
V is the terminal velocity
The Attempt at a Solution
I've tried a few different approaches. I've tried letting u = e^(2g/V) but I never get anywhere by doing this after finding du and substituting that in. My closest answer has been:
[tex]
-V(\frac{e^{-2gt/V} - 1}{e^{-2gt/V} + 1})
[/tex]
and then performing long division to get
[tex]
-V(1 - \frac{2}{1 + e^{-2gt/V}})
[/tex]
When I put that back into the integral I get
[tex]
-V\int dt + 2V\int\frac{dt}{1 + e^{-2gt/V}}
[/tex]
but I don't get the position function and have that negative at the start of the expression... can I get some hints on how to approach this?