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 $t$.
I think you've misread the article. It says rocket engines can attain up to 70% $\eta_c$, which is only the cycle efficiency (how well it turns the energy of the fuel into mechanical energy). This is not the propulsive efficiency.
Unfortunately, for a rocket much of this mechanical energy is used to (wasted..) increase the KE of the exhaust rather than the rocket. As the article mentions, optimum efficiency is when the exhaust speed and rocket speed are matched. But this ends up being horrible for fuel consumption.
Being able to throw the mass of the earth or the atmosphere around makes regular propulsion much more efficient.
In one of your comments you linked to the question Velocity and kinetic energy, violating galilean relativity and said that the efficiency of a car drops with speed. I wouldn't agree with that statement. The question was specifically about interpreting energy in different frames.
If we stick to to just the frame where the ground is at rest (a very valid frame for travel on the earth), then the theoretical efficiency of your battery car approaches 1 as you eliminate drag. The energy of the battery can be given into KE of the vehicle almost entirely since the earth is so massive.
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Hydrogen isn't the only fuel possible, so I presume your question is more general, why is any fuel burned?
You need two things for a rocket: a reaction mass to expel, and a source of power to accelerate it. Combustion rockets combine these two into a single source. The fuel/oxidizer burns generating energy. The energy from combustion heats and then, via the nozzle configuration, accelerates the combustion products as the reaction mass.
Just about anything could be put onboard as the reaction mass, but getting the power to accelerate it is much harder. Batteries and compressed gas hold a bit of energy, but the density is much lower than rocket fuels. Solar panels can gather a nearly unlimited amount of energy, but you have to wait for a long time to collect it. Nuclear fuels could release a lot of power, but putting a nuclear reactor on a rocket takes a lot of mass and is difficult to convince everyone that it can be done safely.
Even if you had sufficient electrical power, converting it into thrust isn't simple. Ion engines can be used, but they have orders of magnitude less thrust than a chemical rocket. The acceleration can be useful in space, but is too small to help lift a rocket off the surface of the earth.
So the fuel is burned because it can be stored on the rocket with a fairly high energy density, and the reaction can take place at a high rate, giving large amounts of thrust.