Not a full answer, but too long for a comment. Maybe this can lead you the right way.
I'm going to ignore the changing mass for a moment and assume a ship with a fuel fraction of $0.5$. If we could get enough thrust from the engines to give a $1g$ thrust to the fuel, we could give a $0.5g$ thrust to the ship.
If you have an engine with $I_{sp} = x$, and a thrust $F$ can burn for $x$ time with a quantity of fuel that has a weight of $F$.
This means engines at that power would run for $100000s$, or just over a day.
Now the nice part is things get better from there. You'd either maintain thrust and increase acceleration to $1g$ as the fuel is exhausted in a day, or you'd throttle down to maintain $0.5g$ and the fuel would last longer. Given that, I assume there's a nice log equation to show the exact relationship between fuel fraction and burn time.
No you can't use mean acceleration in the way you propose, because the equation $t = \sqrt{\frac{2\,s}{a}}$ assumes constant acceleration.
You need to describe the system with a differential equation that takes account of the system's dynamics: since you're learning as a hobby, you may not have seen much of this. Your last paragraph is correct reasoning and is nearer to what you need. The correct wording is "solve" or "invert" or "re-arrange" the equation, but "reverse" is pretty evocative and nearest to "invert".
You need further information to solve your problem: you need a model of how your rocket's mass decreases with time. The simplest (and probably quite accurate model) is that the rate of decrease of mass is some constant mass flow rate: let's call this $q$.
Let's go back to the differential equation whence the Tsiolkovsky equation is derived. We calculate the rocket's change in velocity $\mathrm{d}\,v$ after it has thrown a mass $\mathrm{d}\,m$ out the back at speed $v_e$ relative to it: relative the frame at some instant, before the mass is thrown, the total system's linear momentum is nought: so this must the the momentum relative to this frame after the mass is thrown. The rocket's increase in momentum is $m\,\mathrm{d} v$, which must be balanced by the thrown mass's momentum in the opposite direction so that:
$$m\,\frac{\mathrm{d}v}{\mathrm{d} m} = v_e$$
This is the differential equation which is solved to get the Tsiolkovsky equation. With some juggling, we re-arrange it to:
$$\frac{\mathrm{d}v}{\mathrm{d} t} = \frac{1}{2} \frac{\mathrm{d}v^2}{\mathrm{d} s} = \frac{v_e}{m}\,\frac{\mathrm{d}\,m}{\mathrm{d}t} = \frac{v_e\,q}{m}\tag{1}$$
The first step is a standard identity that converts the acceleration - i.e. the rate of change $\frac{\mathrm{d}v}{\mathrm{d} t}$ of the velocity with respect to time $t$, into a rate of change with respect to the distance travelled $s$. Now, from the Tsiolkovsky equation we have $m(v) = m_0\,\exp\left(-\frac{v-v_0}{v_e}\right)$, where $v_0$ is the beginning velocity and $m_0$ the beginning mass: when we put this into equation (1) we get:
$$\frac{1}{2} \frac{\mathrm{d}v^2}{\mathrm{d} s} = v\,\frac{\mathrm{d}v}{\mathrm{d} s}=\frac{v_e\,q}{m_0}\,\exp\left(\frac{v-v_0}{v_e}\right)\tag{2}$$
This is the differential equation you must integrate to get the distance travelled as a function of $v$. Let me know how you go with this one. Also from (1), we get in the above way from the inverted Tsiolkovsky equation:
$$\frac{\mathrm{d}v}{\mathrm{d} t} = \frac{v_e}{m}\,\frac{\mathrm{d}\,m}{\mathrm{d}t} = \frac{v_e\,q}{m_0}\,\exp\left(\frac{v-v_0}{v_e}\right)\tag{3}$$
which is the differential equation you must solve to get $v$ as a function of time.
Time as a function of distance comes from this last equation. On integrating this last equation, you get
$$v(t) = v_o+v_e\log\left(\frac{m_0}{m_0 - q\, t}\right)$$
and then you need to integrate this, because you now have the differential equation $\frac{\mathrm{d}\,s}{\mathrm{d}\,t}=v_0+ v_e\log\left(\frac{m_0}{m_0 - q\, t}\right)$. This last integration leaves you with:
$$s(t) = v_e\, \left(t-\frac{m_0}{q}\right) \log \left(\frac{m_0}{m_0-q\, t}\right)+t\, (v_0+v_e)$$
To find time to travel a certain distance will need to be done numerically, as, given $s$, you have a transcendental equation in $t$.
Best Answer
Most rocket variants, other than solid rockets (and even some of those) have throttles or some other means of controlling flow/burn rates. Some also have variable Propellant Nozzles, such that there is no one set burn rate. You could perhaps use 'full throttle' or 'max flow rate' if you like, but you have to make that caveat.
No, burn rate is not equivalent to effective exhaust velocity. You can get a rough idea of typical exhaust velocities if you know what you're burning (Kerosene/LOX, Perchlorate/LOX, Hybrid HTPB/N2O, etc...), the rocket velocity, the atmospheric conditions (I.E. earth altitude, vacuum, etc.) and the flow characteristics of the Propellant Nozzle at those conditions. For a rough estimate of Kerosene/LOX, you could start with 4.4km/s.
Now, for what you're after, with a few assumptions (constant acceleration) you can just use the Tsiolkovski + time + Isp of the engine.
Example: I have 4000 kg of a fuel in a 8000 kg fully fueled rocket in space, the fuel is of a certain composition such that through my nozzle design at a vacuum it exits my craft at effectively 4.0km/s ($I_{sp}$), and at full throttle it would burn all it's fuel in 10 seconds. Using the Tsiolkovski:
$ \Delta V = v_e * ln(\frac {m_0} {m_1} ) $
$ \Delta V = 4.0 km/s * ln (\frac {8000} {4000} ) $
$ \Delta V = 4.0 km/s * ln (2) $
$ \Delta V = 4.0 km/s * 0.693... = 2.77 km/s $
This change in velocity was made over 10 seconds, so
$ a = \frac {\Delta V\ km/s} {10\ s} = \frac { 2.77\ km/s } {10\ s} = 0.277\ km/s^2 = 277\ m/s^2 $