As Kwaśnicki remarked, the velocity process $v_t$ is a *Brownian bridge*  $$
v_t = v_0 (1 - \frac{t}{T}) + v_T \frac{t}{T} + (T - t) \int_0^t \frac{1}{T-s} d B_s \;.
$$ This is a Gaussian process on the interval $[0,T]$ with initial value $v_0$ and final value $v_T$.  

As before, the position process $x_t$ is obtained by integrating the velocity process: \begin{align*}
x_T &= x_0 + \int_0^T v_t dt  \\
&=x_0 + \frac{T}{2} (v_0 + v_T) + \int_0^T \int_0^t   \frac{T-t}{T-s} d B_s dt \\ 
&=x_0 + \frac{T}{2} (v_0 + v_T) + \int_0^T \left( \int_s^T   \frac{T-t}{T-s} dt \right) d B_s 
\end{align*}
This double integral is Gaussian with mean zero and variance $T^3 / 12$.  Hence, $$
x_T \sim \mathcal{N}( x_0 + \frac{T}{2} (v_0 + v_T), \frac{T^3}{12} )
$$