What does Power really mean? I've been trying to solve a problem for some time. I have been given conflicting information by both literature, colleagues and people on this very forum.
It's a very simple question: What is the definition of power?
In particular, I would like to know how constant power would accelerate a constant mass over a fixed distance, assuming that no resistive forces are at work, i.e. no friction, no air resistance, etc.
I know that power is the rate at which work is done with respect to time, i.e. $P = \operatorname{d}\!W/\operatorname{d}\!t$. 
One formula that I have seen is that $P = Fv$, i.e. power is the product of force and velocity. According to Wikipedia, this requires $v$ to be constant. Someone on this site, and someone I work with have both said that I need $v$ to be constant for $P=Fv$ to hold. 
However, I found an article from 1930, which found equations of motion for constant power. Just like the standard equations of motion $v=u+at$, $v^2=u^2+2as$, $s=ut+\frac{1}{2}at^2$, and $s=\frac{1}{2}(u+v)t$ all assume a constant force, i.e. a constant acceleration, L. W. Taylor (1930) found six analogous formulae for motion under constant power. He used $P= Fv$ as his definition for power.
Can I work with $P=Fv$, assuming that $F$ and $v$ vary, provided their product is constant?
 A: The equation $P = Fv$ always holds, even when neither the force, the velocity nor their product are constant. If you write:
$$ P(t) = F(t)v(t) $$
Then this is true at all times $t$ and gives the instantaneous power at the time $t$. Obviously if the velocity and/or force vary with time then the power will also vary with time.
A: There is a precise definition of power in the following question: Why isn't the product rule used in the definition of mechanical work?.  Here is a reproduction of that text:

For a particle traveling along a parameterized curve $\vec x(t)$ and under the influence of a force $\vec F(\vec x,t)$ which is explicitly dependent on both position in space and time, the work performed on the particle by this force from a time $t_0$ to a time $t$ is defined as follows:
  \begin{align}
  W_{t_0}(t) = \int_{t_0}^{t} \vec F(\vec x(t'),t')\cdot \dot{\vec x}(t') \,dt'
\end{align}
  Note that the expression on the right is often written $\int \vec F\cdot \vec dx$, but this is really schematic, the the mathematically precise definition is what I have written above in terms of a parameterized path with an integral over some range of parameter values.  The definition of instantaneous power is then
  \begin{align}
  P(t) = \dot W_{t_0}(t)
\end{align}
  Taking the derivative of both sides with respect to $t$, and using the fundamental theorem of calculus, we obtain the desired expression for the power
  \begin{align}
  P(t) = \vec F(\vec x(t),t)\cdot \dot{\vec x}(t)
\end{align}

Notice that the expression $\vec F\cdot \vec v$ is the instantaneous power for arbitrarily changing force and velocity.  There is not even a restriction on the constancy of their product.
