= Solution
Put $\omega_0=\sqrt{k/m}$ and $\gamma=b/(2m)$. The characteristic roots are $-\gamma\pm\sqrt{\gamma^2-\omega_0^2}$. For release from rest at <displacement> $x_0>0$, the three regimes are
$$
x(t)=
\begin{cases}
x_0e^{-\gamma t}\left[\cos(\omega_dt)+\dfrac{\gamma}{\omega_d}\sin(\omega_dt)\right],
&0<\gamma<\omega_0,\quad \omega_d=\sqrt{\omega_0^2-\gamma^2},\\
x_0(1+\omega_0t)e^{-\omega_0t},
&\gamma=\omega_0,\\
x_0e^{-\gamma t}\left[\cosh(st)+\dfrac{\gamma}{s}\sinh(st)\right],
&\gamma>\omega_0,\quad s=\sqrt{\gamma^2-\omega_0^2}.
\end{cases}
$$
<underdamping> produces decaying oscillations; <critical damping> is the repeated-root boundary; <overdamping> has two real decay rates. The critical and overdamped curves from rest approach equilibrium without crossing it. All three initially have zero slope and negative <acceleration>. A sufficiently strong initial kick can cause a single crossing even without an oscillatory mode.
For the critical oscillator with initial <velocity> $-v_0$, the repeated-root solution is
$$
x(t)=\left[x_0+(\omega_0x_0-v_0)t\right]e^{-\omega_0t}.
$$
Its exponential is positive at finite time. Therefore the bracket has a positive-time zero precisely when
$$
\boxed{v_0>\omega_0x_0,\qquad
t_{\rm cross}=\frac{x_0}{v_0-\omega_0x_0}.}
$$
Equality gives a pure positive exponential and no finite-time crossing.
For the specified numerical oscillator, $\omega_0=1\ {\rm s^{-1}}$ and $\gamma=2/\sqrt3\ {\rm s^{-1}}>1\ {\rm s^{-1}}$, verifying <overdamping>. Its two roots are $-1/\sqrt3$ and $-\sqrt3$ in inverse seconds. The initial conditions give
$$
\boxed{x(t)=
(0.015-0.015\sqrt3)e^{-t/\sqrt3}
+(0.015\sqrt3-0.005)e^{-\sqrt3t}\quad{\rm m},}
$$
with $t$ in seconds. The slow component has a negative coefficient; the fast positive component initially dominates.
The values for the requested plot are
$$
\begin{array}{c|rrrrrrr}
t\ ({\rm s})&0&0.5&1.0&1.5&2.0&2.5&3.0\\
x\ ({\rm mm})&10.000&0.598&-2.452&-3.057&-2.804&-2.317&-1.827
\end{array}
$$
The curve crosses near $0.56\ {\rm s}$ and then returns toward zero from below. The exact expression provides a check on the graphical estimate:
$$
\boxed{t_{\rm cross}=\frac{\sqrt3}{2}
\ln\!\left(\frac{0.015\sqrt3-0.005}{0.015\sqrt3-0.015}\right)
\simeq0.561\ {\rm s}.}
$$
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2016/paper-2-damping-and-guidance.png]
{title=Original damped-oscillator solutions and clamped-sheet waveguide calculations}
{description=Rest-release curves compare three <damping> regimes; the separate kicked overdamped trajectory includes the requested sampling times. The guided-mode panel plots the dispersion branches, and the profile panel compares the two-mode input with its first phase-reversed transverse profile.}
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