As Kwaśnicki remarked, the velocity process $v_t$ is a *Brownian bridge*, which can be represented as:  $$
v_t = v_0 (1 - \frac{t}{T}) + v_T \frac{t}{T} + (T - t) \int_0^t \frac{1}{T-s} d B_s \;.
$$ (For an intro to this representation, see the first exercise of the following [exercise sheet on Brownian bridges][2]). 

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, $$
x_T \sim \mathcal{N}( x_0 + \frac{T}{2} (v_0 + v_T), \frac{T^3}{12} ) \;.
$$
Note that the variance of this process with pinned initial and final velocities is a quarter of the variance of the position process with pinned initial and unpinned final velocity -- which makes sense intuitively. 


  [1]: https://books.google.com/books?id=zidqvu_O73cC&pg=PA362&lpg=PA362&dq=brownian%20bridge%20pathwise%20representation&source=bl&ots=7D53aWPAZZ&sig=b_RJuzmS4hhur-vDJyTOrl8W-_g&hl=en&sa=X&ved=0ahUKEwiCy9W1j9rYAhVMMd8KHQ_ED0sQ6AEIbDAI#v=onepage&q=brownian%20bridge%20pathwise%20representation&f=false
  [2]: http://wt.iam.uni-bonn.de/fileadmin/WT/Inhalt/people/Karl-Theodor_Sturm/Lectures/vorlesungSS09/sheet01.SS09.pdf