Can inverse velocities add? Everyone here probably saw once the question of a container being filled with multiple hoses each at it's own speed, or maybe lumberjacks working or machines or whatever, the question usually goes something like the following:
If $thing_1$ takes $x$ seconds to do something and $thing_2$ takes $y$ seconds then how much time it takes for them to do it together? The answer is $$(1/x+1/y)^{-1}=xy/(x+y).$$
My problem is that the way I solve this in my head is that I first convert $x,y\frac{(seconds)}{(operation)}$ to $1/x,1/y\frac{(operations)}{(second)}$ then I add them and invert them again to get the total new seconds per operation, but why can't I add them just straight, I understand that I would be adding how much operations are done in a second, and adding them would be like saying if you have one second then how many operations you're doing by joining the "things", but it still should be giving me the correct amount of operations if the units are additive. I.e: "If one lumberjack passes one hour cutting a tree ($1\frac{hour}{tree}$) then two lumberjacks pass two hours cutting one tree? ($2\frac{hour}{tree}$, obviously wrong)"
 A: The answer has to do with algebra, if you think of each quantity as a ratio (fraction).
If you have a common denominator in algebra then the addition is simple
$$ \frac{x}{A} + \frac{y}{A} = \frac{x+y}{A} $$
you just add up the quantities and place keep them in the numerator or $x \oplus y \rightarrow x+ y$.
But if you have a common numerator and different denominators, then you need to do the harmonic addition (as its called)
$$ \frac{A}{x} + \frac{A}{y} = \frac{A y}{x y} + \frac{A x}{x y} = \frac{A y + A x}{x y} = \frac{A} {\left( \frac{x y}{x + y} \right) }$$
And now the denominators add as $x \oplus y \rightarrow \frac{x y}{x+y}$
In the cases mentioned you are dealing with velocities, which are a ratio of amount and time. The amount is fixed and time is what you are trying to add, which as you understand is in the denominator. So the harmonic addition rules apply.
A: John Alexiou's answer is best. But we can think about it as you were.
If one lumberjack's rate is $1 \frac{hour}{tree}$, it is also $\frac{1/2 \space hour}{1/2 \space tree}$.
Two lumberjacks can cut both $1/2$ trees in the same $1/2$ hour, so their combined rate is $\frac{1/2 \space hour}{1 \space tree}$
