= Solution
For a <Lorentz transformation> to an <inertial frame> moving at speed $v_B$ in the positive $x$ direction, write $\beta_B=v_B/c$ and $\Gamma_B=(1-\beta_B^2)^{-1/2}$. Any <four-vector> transforms as
$$
\boxed{W'^0=\Gamma_B(W^0-\beta_BW^1),\quad W'^1=\Gamma_B(W^1-\beta_BW^0),\quad W'^2=W^2,\quad W'^3=W^3.}
$$
This convention applies to $X=(ct,x,0,0)$ and uses the same $+---$ <Minkowski metric> as the <four-momentum> calculation.
The definition of <proper time> gives $d\tau=dt\sqrt{1-u^2/c^2}=dt/\gamma$ for $|u|<c$. Hence the <four-velocity> is
$$
\boxed{U=\gamma(c,u).}
$$
Since $\dot\gamma=\gamma^3u\dot u/c^2$, differentiating with respect to <proper time> gives
$$
A=\gamma\frac{dU}{dt}=\left(\frac{\gamma^4u\dot u}{c},\gamma^2\dot u+\frac{\gamma^4u^2\dot u}{c^2}\right)=\boxed{\gamma^4\dot u\,(u/c,1)}.
$$
Here $1+\gamma^2u^2/c^2=\gamma^2$ simplifies the spatial component of the <four-acceleration>.
At each event, boost with $v_B=u$ to the instantaneous rest <inertial frame>. Then $A'^0=0$ and $A'^1=\gamma\gamma^4\dot u(1-u^2/c^2)=\gamma^3\dot u$. Thus the signed <proper acceleration for one-dimensional motion> is
$$
\boxed{a=\gamma^3\dot u=\frac d{dt}(\gamma u).}
$$
The last equality follows by differentiating $\gamma u$; its magnitude is the usual scalar <proper acceleration>.
If this signed <proper acceleration> is constant and the particle starts from rest, integration gives $\gamma u=at$, so
$$
u(t)=\frac{at}{\sqrt{1+a^2t^2/c^2}}.
$$
Integrating once more and imposing $x(0)=0$ yields the <hyperbolic motion> worldline
$$
\boxed{x(t)=\frac{c^2}{a}\left(\sqrt{1+\frac{a^2t^2}{c^2}}-1\right)\quad(a\ne0).}
$$
For $a=0$ the continuous limiting solution is $x=0$. Finally, the <Taylor series> of the square root gives
$$
x(t)=\frac12at^2-\frac{a^3t^4}{8c^2}+O(a^5t^6/c^4),\qquad \boxed{x\sim\frac12at^2\quad\text{when }|at|\ll c.}
$$
This recovers constant-acceleration <Newtonian mechanics> at low speed while the exact relativistic speed remains less than $c$.
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