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 $$
x_t \sim \mathcal{N}( x_0 + t v_0, \frac{t^3}{3}) \;.
$$


----------

> **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][1], $$
\frac{S_N}{s_n} \overset{d}{\to} \mathcal{N}(0,1) \;,
$$  from which it follows that, $$
X_N \overset{d}{\to}  \mathcal{N}(x_0 + t v_0, \frac{t^3}{3} ) \;.
$$


  [1]: https://en.wikipedia.org/wiki/Central_limit_theorem#Lyapunov_CLT