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Merbdon
Merbdon is offline
#5
Jan4-12, 02:18 PM
P: 4
Thanks for the replies everyone, but I'm still not seeing it. I am not trying to copy and paste an answer here, I'm doing this on my own time, trying to study before I go back to school in the Fall.
Where does the e come from? Is there some basic calculus that I'm overlooking? It looks to me that solving the problem should go something like this:

[itex]\frac{dv}{dt}[/itex]= g - kv

∫[itex]^{t}_{0}[/itex]dvdt = ∫[itex]^{t}_{0}[/itex](g - kvt)dt

v = gt - kvt → v = [itex]\frac{gt}{1 + kt}[/itex]

Which makes sense, but doesn't match the answer in the book. Is the book incorrect, or am I? What am I missing here?

Thanks again for the help everyone.
-Merb