Something I'd like to clear up. Calculating the 4-acceleration from the 4-velocity is a very straightforwards procedure that can be found in most texts, in the context of either special relativity or general relativity.
The tricky part of the problem problem (IMO) is finding the 4-velocity of Superman, aka the sliding block. I do this in
https://www.physicsforums.com/threads/the-notion-of-weight-in-relativity.701257/#post-4445483, but it's helpful to eliminate the factor K I used in the derivation, and replace it with ##\beta_0##, the block's normalized velocity, and ##\gamma_0 = 1/\sqrt{1-\beta_0^2}##
The formulae to do the conversion are in the original post, but the converted results are not written down explicitly. Doing so now, we find
$$\frac{dt}{d\tau} = \gamma_0 \sqrt{1+g^2\,t^2} \quad \frac{dx}{d\tau} = \beta_0 \, \gamma_0 \quad \frac{dz}{d\tau} = \gamma_0 \, g \, t$$
This express the 4-velocity in terms of t, not ##\tau##. My original intent was to stop here, but as I wrote this I realized it'd be helpful to go further and additionaly re-write the proper velocity as a function of ##\tau## rather than t. To do this we make use of the relation between t and ##\tau## derived in the original post:
$$\tau = \frac{\sinh^{-1} gt}{ \gamma_0 g} \quad t = \frac{1}{g} \sinh \gamma_0 g \tau $$
We can then substitute to find the 4-velocity as a function of ##\tau##
$$\frac{dt}{d\tau} = \gamma_0 \sqrt{1+\sinh^2 \gamma_0 g \tau} = \gamma_0 \cosh \gamma_0 g \tau \quad \frac{dx}{d\tau} = \beta_0 \gamma_0 \quad \frac{dz}{d\tau} = \gamma_0 \, \sinh \gamma_0 g \tau $$
While I'm at, I realized I can write the 3-velocity of Superman as a function of time:
$$\frac{dx}{dt} = \frac{\beta_0}{\sqrt{1+g^2\,t^2 }} \quad \frac{dz}{dt}= \frac{g\,t}{ \sqrt{1+g^2t^2}} $$
We can see that the velocity ##dz/dt## is the same function of time t as for the
relativistic rocket, as it should be.
