The position process $x_t$ satisfies $$ x_t = x_0 + t v_0 + \int_0^t W_s ds \;. $$ Because $\int_0^t W_s ds \sim \mathcal{N}(0,\frac{1}{3} t^3)$, a simple change of variables shows that $$ x_t \sim \mathcal{N}( x_0 + t v_0, \frac{t^3}{3}) \;. $$
The following relates $x_t$ to a discrete-time weak approximation $X_N$, which is obtained by weakly approximating the Brownian increments in the velocity process $v_t$ by Rademacher random variables.
Claim. Given $t>0$ and $N \in \mathbb{N}$, set $$ \tag{$\star$} X_n = X_{n-1} + h V_n \;, \quad V_n = V_{n-1} + \sqrt{h} A_n \;, \quad 1 \le n \le N $$ with initial conditions $X_0 = x_0$, $V_0 = v_0$, and where $\{ A_i \}_{i=1}^N$ are independent Rademacher random variables and $h=\frac{t}{N}$. For any $t>0$, $$ X_{N} \overset{d}{\to} x_t \quad \text{as $N \to \infty$} \;. $$
Proof. Unraveling the recurrence relation ($\star$) yields, $$ X_N = x_0 + t v_0 + \frac{t^{3/2}}{N^{1/2}} S_N $$ where $S_N = \sum_{i=1}^N A_i \frac{(N-i+1)}{N}$. Note that $S_N$ is a sum of independent and uniformly bounded random variables each with mean zero. Moreover, the variance $s_N^2$ of $S_N$ goes to $\infty$ as $N \to \infty$ since $$ s_N^2 = \operatorname{Var}(S_N) = \frac{1+3 N + 2 N^2}{6 N} \;. $$ By the Lyapunov Central Limit Theorem, $$ \frac{S_N}{s_n} \overset{d}{\to} \mathcal{N}(0,1) \;, $$ from which it follows that, $$ X_N \overset{d}{\to} \mathcal{N}(x_0 + t v_0, \frac{t^3}{3} ) \;. $$