= Solution
Take $x(0)=0$. Integrating the <velocity> solution gives
$$
x(t)=v_0\tau(1-e^{-t/\tau})+\frac1\zeta\int_0^t[1-e^{-(t-s)/\tau}]F(s)\,ds.
$$
The initial/noise cross term vanishes under the independence assumption. Evaluating the remaining integral yields the general <integrated Ornstein-Uhlenbeck displacement>
$$
\langle x(t)^2\rangle=\langle v_0^2\rangle\tau^2(1-e^{-t/\tau})^2+\frac A{\zeta^2}\left[t-2\tau(1-e^{-t/\tau})+\frac\tau2(1-e^{-2t/\tau})\right].
$$
Use the thermal initial <velocity> and the <fluctuation-dissipation relation for a Langevin particle>. With <diffusion coefficient> $D=k_BT/\zeta$, this simplifies to
$$
\boxed{\langle x(t)^2\rangle=2D\left[t-\tau(1-e^{-t/\tau})\right].}
$$
For $t\ll\tau$, expansion gives $\langle x^2\rangle=(k_BT/m)t^2+O(t^3)$: displacement is initially $v_0t$, so the particle is \b[ballistic]. For $t\gg\tau$, $\langle x^2\rangle=2Dt-2D\tau+o(1)$, hence the leading behavior is \b[one-dimensional <diffusion> with $D=k_BT/(6\pi\mu a)$].
The ballistic conclusion depends on the initial state. For a particle prepared exactly at rest, $v_0=0$, the first expression gives $\langle x^2\rangle=A t^3/(3m^2)+O(t^4)$ at very short times. A thermal initial <velocity>, or a nonzero initial <velocity> variance, is required for the stated $t^2$ mean-square law. It should not be replaced silently by a zero-<velocity> initial condition.
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