Almost all paths $t\mapsto B_t(\omega)$ of the <PERSON> process are in fact not differentiable. Therefore, if we take $B_t$ as position of the particle $x(t)$ we end up having an unphysical particle whose velocity is not well defined. As far as I know, this led <PERSON>, <PERSON> and <PERSON> to develop a dynamical theory of Brownian motion by which the velocity $v(t)=\dot x(t)$ is 'driven' by a <PERSON> process $$ v(t)=e^{-\beta t}v_0+e^{-\beta t}\int_0^te^{\beta s}\,dB_s $$ which is described in <PERSON> [1]. The parameter $\beta$ describes friction. The larger that is the faster the velocity tends to zero. If $\beta=0$ we have $$ v(t)=v_0+B_t\,,\quad\quad x(t)=x_0+\int_0^tv(s)\,ds=x_0+v_0t+\int_0^tB_s\,ds\,. $$ In this theory the mean (over many particles) of $x(t)v(t)$ is \begin{align} \mathbb E[x(t)v(t)]&=\mathbb E\left[B_t\int_0^tB_s\,ds\right]=\mathbb E\left[B_t\int_0^t(t-s)\,dB_s\right]=\int_0^t(t-s)\,ds=t^2/2. \end{align} This is a covariance. The correlation turns out to be $$ \frac{\mathbb E[x(t)v(t)]}{\sqrt{\mathbb E[x^2(t)]}\sqrt{\mathbb E[v^2(t)]}}=\frac{1}{2\sqrt{3}}\,. 649
$$ Note that the Langevin/OU theory is not compatible with $\langle x^2\rangle=2Dt$. That equation is due to <PERSON> and <PERSON> which <PERSON> calls a highly idealized treatment in his Chapter 9. In fact in <PERSON>'s notation, <PERSON> Smoluchovski corresponds to $$ x(t)=\sqrt{2D}B_t $$ which <PERSON>/OU discarded as unphysical. <PERSON> also writes that the predictions of the OU theory are numerically indistinguishable of the <PERSON> <PERSON> theory.
I also want to mention the approach using white noise by which the derivative of the <PERSON> process exists in the sense of distributions (delta functions): $$ w(t)=\dot B_t=\frac{dB_t}{dt}\,,\quad\quad E[w(t)w(s)]=\delta(t-s)\,. $$ Informally we can write this as $$ \mathbb E[dB_t\,dB_s]=\delta(t-s)\,dt\,ds $$ which (informally) yields $$ \mathbb E[dB_t\,B_t]=\int_0^t\delta(t-s)\,dt\,ds=dt\,. $$ Therefore, the covariance is $$ \mathbb E[\dot B_tB_t]=1\,. $$ In this theory we could set the position $x(t)$ to be $B_t$ but, of course, have the unphysical velocity $v(t)=\dot B_t$ (delta function). Note that the variance $\mathbb E[(\dot B_t)^]$ is infinite by the definition of white noise. I could not figure out quickly if the mathematicians therefore define the correlation of $\dot B_t$ and $B_t$ to be zero or if the let it be undefined.
[1] <PERSON>, Dynamical Theories of Brownian Motion 346