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 $t$.