 Solution

Let $f=|\mathbf{f}|$ and $\hat{\mathbf{f}}=\mathbf{f}/f$. A constant laboratory three-force means $d\mathbf{p}/dt=\mathbf{f}$, hence $\mathbf{p}=\mathbf{f}t$ from rest. Using $E^2=m^2{c}^4+p^2{c}^2$ givesThe dimensionless time $s=ft/(mc)$ makes the shapes transparent: $\gamma =\sqrt{1+s^2}$ starts at one with zero slope and becomes asymptotically linear; $\beta =s/\sqrt{1+s^2}$ starts linearly and approaches one without reaching it at finite time. The acceleration decreases as $f/(m{\gamma }^3)$.

Integrating the velocity gives the distance along the force,Thus a constant three-force produces hyperbolic motion with constant proper acceleration $f/m$, although its velocity is never Newtonian at large time. The following original curves supply the requested sketches.