I'm not at all confused by the chain rule, I'm confused by how to employ it from the starting conditions.
I've successfully solved the equation, which looks like 1/k*ln[(cosh(1/k*arctanh(v/√gk))] evaluated from v0 to v1
I cannot imagine this is what the prof is looking for though (even though...
How do you derive dv/dx and dx/dt from equations with neither of these terms?
ie.
dv/dt = g-mv
dy/dt grabbing the already solved equation (y would be the falling problem)
yields
dy/dt = \sqrt{g/k}tanh(\sqrt{g/k}t
dv/dy = 0?
I see no way to convert unless I'm missing key steps here
if you sub...
Here's what I get:
t + C = 1/2*\sqrt{k/g}[ln\frac{v-√g/k}{v+√g/k}]
When I work out the rest nothing simplifies.
I've also tried it using your first suggestion stuckAgain.
Using some info in this book (taylor classical)
I got v = \sqrt{g/k} tanh(\sqrt{g}t/\sqrt{k}) but this is the general equation?
Homework Statement
A particle of mass "m" whose motion start with downard velocity V0 in a constant gravitational field. The drag force is quadratic and proportional to kmv2. What is the distance s through which the particle falls in accelerating from v0 to v1. Give your expression for s in...