= Solution
Let $m$ denote the constant <rest mass>, set $x(0)=0$, and put $a=f_x/m$. The relativistic three-<force> is $f_x=dp_x/dt$, so
$$
p_x=f_xt,\qquad p_x=\gamma m\dot x,\qquad
\gamma=\sqrt{1+\frac{f_x^2t^2}{m^2c^2}}.
$$
It follows that
$$
\boxed{\dot x(t)=\frac{at}{\sqrt{1+(at/c)^2}},\qquad
\ddot x(t)=\frac{a}{[1+(at/c)^2]^{3/2}}.}
$$
Integrating the <velocity> with the initial position fixed gives
$$
\boxed{x(t)=\frac{c^2}{a}\left[\sqrt{1+(at/c)^2}-1\right].}
$$
These formulas allow either sign of $a$; their $a\to0$ limit is rest at the origin. For $a\ne0$, the trajectory obeys
$$
\left(x+\frac{c^2}{a}\right)^2-c^2t^2=\frac{c^4}{a^2}.
$$
This is <hyperbolic motion>. Its coordinate <acceleration> decreases while its <proper acceleration> has constant magnitude $|a|$. In particular, a constant three-<force> does not give constant coordinate <acceleration> at relativistic <speed>.
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