Calculate acceleration from power I am currently programming a simple computer space ship simulation. This is just for fun to train my programming, 3D graphics and a little bit of physics skills. What I want is the user to pick a space ship that has a mass and power. I could just use force, but somehow power sounds better, because you can compare it to cars if you have no knowledge in physics. Without any reason I would like that to have at least a small physics background that is not completely off. So here is my question:
What is the acceleration of a 75 kg space ship with a 1 hp rocket engine?
I know there is nothing like a 1 hp rocket engine, but let's define:
A 1 hp rocket engine is an engine that has the power to raise a 75 kg space ship against the earth's gravitational force of $9.80665 \, \text{m/s}^2$ over a distance of one metre in one second (see Wikipedia).
Now let's assume there is no air friction and the space ship somehow does not lose mass while accelerating. This does not have to be completely realistic. What is the acceleration of the space ship, if you turn off the earth's gravitational force?
I have 
$$P = \frac{W}{t} = \frac{F\cdot s}{t} = m a v$$
so
$$ a = \frac{P}{m v}$$
but I do not know the velocity $v$.
I hope you can somehow point me in the right direction or provide a better solution for my simulation.
 A: If you really had a "constant power" engine, and all that power was transferred to your rocket which does not lose mass, it would result in a linear increase in the kinetic energy.
And since the kinetic energy $E=\frac12 m v^2$, you can find the velocity at a given time from
$$P\cdot t = \frac12 m v^2\\
v = \sqrt{\frac{2 \cdot P \cdot t}{m}}$$
If you wanted the acceleration as a function of time, you would differentiate...
$$a = \frac{dv}{dt} = \sqrt{\frac{ P }{2m\cdot t}}$$
But note


*

*Rocket engines provide thrust not power

*Mass changes significantly as fuel is used up


So this is totally unrealistic.
A: As @Floris said, rocket motors provide thrust (reaction force), not power. The thrust is just the mass times velocity of exhaust, per second. The power is the mass times velocity squared, per second (over two).
Example, shooting a rifle bullet has low reaction force, but high energy. Shooting a bowling ball with the same reaction force takes a lot less energy.
So if a rocket engine could eject bowling balls, it would take a lot less energy than if it could eject something very light (at very high speed).
So bowling balls would make excellent exhaust mass.
The trouble is, you'd run out of them very quickly.
The reason rockets have very high power is just so they can make the fuel last as long as possible.
That's why ion-engines are being considered for long-term thrusting in space.
They sip fuel by ejecting it at a velocity far higher than a chemical reaction could generate.
A: 
What is the acceleration of the space ship, if you turn off the earth's gravitational force?

The acceleration of the ship is
$$
a = \frac{F + G}{m}
$$
where $F$ is thrust of the engine (force), $m$ is mass of the rocket and $G=-mg$ is gravity force. If thrust is such as to exactly cancel gravity, it is $F=mg$ so the acceleration with no gravity would be $g$.
In your question, however, you mention power instead of thrust. From your formulation, I assume you mean part of energy liberated from the fuel per unit time that turns into kinetic energy of the rocket. Unfortunately, this number changes as the rocket accelerates, because it is equal to
$$
P = F.v
$$
where $v$ is velocity of the rocket. $F$ is constant, but as $v$ changes with time, $P$ changes with time as well. When the rocket starts, $v = 0$ so $P=0$. The faster the rocket moves, the higher the power $P$ is. Quantity $P$ is thus not good for characterizing the capability of the rocket engine (or jet engine) alone.
While the efficient power $P$ changes in time, total power liberated $Q$ (including the energy leaving with exhausts) may be constant. It can be written as
$$
Q = B.\epsilon
$$
where $\epsilon$ is energy liberated per unit mass of exhaust gas and $B$ is mass loss per unit time (rate of fuel consumption). If $B$ is constant, so is $Q$.
From the law of conservation of momentum, thrust can be expressed as
$$
F = B.u,
$$
where $u$ is speed of the gas exhausts leaving the nozzle of the rocket in the frame of the engine. It is constant in time, a characteristic of the fuel mixture and the engine.
Combining the last two relations, we obtain
$$
F = Q \frac{u}{\epsilon}
$$
and the acceleration is thus given by 
$$
a = \frac{Qu}{m\epsilon}.
$$
Thus, all other things being equal, acceleration would be proportional to the liberated power  $Q$. In practice, changing $Q$ requires changing other things as well, which will affect $u$ and if mixture ratio changes, even $\epsilon$. Thus it is not necessarily true that bigger engine with higher $Q$ will have proportionally higher $a$.
