= 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 $E^2=m^2c^4+p^2c^2$ gives
$$
\boxed{\gamma(t)=\sqrt{1+\left(\frac{ft}{mc}\right)^2},\qquad
\boldsymbol\beta(t)=\frac{ft/(mc)}{\sqrt{1+[ft/(mc)]^2}}\,\hat{\mathbf f}.}
$$
The 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,
$$
\ell(t)=\frac{mc^2}{f}[\gamma(t)-1].
$$
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.
\Image[/past-exam-of-the-symmetry-and-relativity-course-of-the-university-of-oxford/2016-constant-force-kinematics.png]
{title=Lorentz factor, speed and world line for acceleration from rest under a constant laboratory three-force}
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