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, $$ x_T \sim \mathcal{N}( x_0 + \frac{T}{2} (v_0 + v_T), \frac{T^3}{12} ) $$

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).

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.