= Solution
With $\gamma=(1-v^2/c^2)^{-1/2}$,
$$
U^\mu=\gamma(c,v),\qquad U_\mu U^\mu=\gamma^2(c^2-v^2)=c^2.
$$
<Differentiation> with respect to <proper time> gives
$$
A^\mu=\left(\frac{\gamma^4v\cdot a}{c},\ \gamma^2a+\frac{\gamma^4(v\cdot a)v}{c^2}\right).
$$
Orthogonality $U\cdot A=0$ implies that in the instantaneous rest frame $A'^\mu=(0,a')$. For collinear motion a Lorentz boost gives
$$
a'=\gamma^3a=\frac{a}{(1-v^2/c^2)^{3/2}}.
$$
For constant positive <proper acceleration> $a'$,
$$
v(t)=\frac{a't}{\sqrt{1+(a't/c)^2}},\qquad
x(t)=\frac{c^2}{a'}\left(\sqrt{1+(a't/c)^2}-1\right).
$$
This hyperbola has future null asymptote $x=ct-c^2/a'$. A right-moving signal emitted from $x_0$ at $t=0$ follows $x=ct+x_0$; if $x_0<-c^2/a'$, it lies beyond that <acceleration> horizon and never intersects the worldline.
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