 Solution

Let $m$ denote the constant rest mass, set $x(0)=0$, and put $a=f_x/m$. The relativistic three-force is $f_x=d{p}_x/dt$, soIt follows thatIntegrating the velocity with the initial position fixed givesThese formulas allow either sign of $a$; their $a\to 0$ limit is rest at the origin. For $a\ne 0$, the trajectory obeysThis 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.