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 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: $\frac{dv}{dt}$= g - kv ∫$^{t}_{0}$dvdt = ∫$^{t}_{0}$(g - kvt)dt v = gt - kvt → v = $\frac{gt}{1 + kt}$ 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