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We know the covariance $E[\int_0^tW(u)\,du\int_0^tW(v)\,dv]=t^3/3\,.$ It is easy to see that the covariance between position $x(t)=\int_0^tW(u)\,du$ and velocity $v(t)=\dot x(t)=W(t)$ is $E[(\int_0^tW(u)\,du) W(t)]=t^2/2\,.$ Therefore, the correlation between $x$ and $v$ is

$$ \varrho=\frac{t^2/2}{\sqrt{t^3/3}\sqrt{t}}=\frac{\sqrt{3}}{2}\,. $$ Let's normalize the variables to make them standard normal: $$ \hat{x}(t)=\frac{x(t)}{\sqrt{t^3/3}}\,,~~\hat{v}(t)=\frac{v(t)}{\sqrt{t}}\,. $$ Then, $P(\hat x,\hat v,t)$ is the density of a bivariate normal distribution with correlation $\varrho\,$: $$ P(\hat x,\hat v,t)=\frac{1}{2\pi\sqrt{1-\varrho^2}}\,\exp\Big(-\frac{\hat x^2-2\varrho\,\hat x\,\hat v+\hat v^2}{2(1-\varrho^2)}\Big)\,. $$ Therefore, \begin{eqnarray}\label{ePDF} P(x,v,t)&=&\frac{\sqrt{3}}{2\pi t^2\sqrt{1-\varrho^2}}\,\exp\Big(-\frac{3x^2/t^3-2\sqrt{3}\varrho\,x\,v/t^2+v^2/t}{2(1-\varrho^2)}\Big)\\ &=&\frac{\sqrt{3}}{\pi t^2}\,\exp\Big(-\frac{6x^2-6\,x\,v\,t+2v^2\,t^2}{t^3}\Big)\,. \end{eqnarray} This function satisfies the Kolmogorov PDE $$ \partial_t P=-v\,\partial_x P+\frac{1}{2}\partial_{vv}P $$ for the diffusion process $$ \left(\begin{array}{c}dx\\dv\end{array}\right)=\left(\begin{array}{c}v\\0\end{array}\right)\,dt+\left(\begin{array}{cc}0&0\\1&0\end{array}\right)\left(\begin{array}{c}dW\\dW_v\end{array}\right) $$ where $W_v$ is a dummy BM that is not driving anything.

Regarding the joint distribution of position and velocity:

From my previous answer we know the variance $E[(\int_0^tW(u)\,du)^2]=t^3/3\,$ of the position. It is easy to see that the covariance between position $x(t)=\int_0^tW(u)\,du$ and velocity $v(t)=\dot x(t)=W(t)$ is $$ E[\textstyle(\int_0^tW(u)\,du) W(t)]=t^2/2\,. $$ Therefore, the correlation between $x$ and $v$ is

$$ \varrho=\frac{t^2/2}{\sqrt{t^3/3}\sqrt{t}}=\frac{\sqrt{3}}{2}\,. $$ Let's normalize the variables to make them standard normal: $$ \hat{x}(t)=\frac{x(t)}{\sqrt{t^3/3}}\,,~~\hat{v}(t)=\frac{v(t)}{\sqrt{t}}\,. $$ Then, $P(\hat x,\hat v,t)$ is the density of a bivariate normal distribution with correlation $\varrho\,$: $$ P(\hat x,\hat v,t)=\frac{1}{2\pi\sqrt{1-\varrho^2}}\,\exp\Big(-\frac{\hat x^2-2\varrho\,\hat x\,\hat v+\hat v^2}{2(1-\varrho^2)}\Big)\,. $$ Therefore, the joint PDF of position and velocity is \begin{eqnarray}\label{ePDF} P(x,v,t)&=&\frac{\sqrt{3}}{2\pi t^2\sqrt{1-\varrho^2}}\,\exp\Big(-\frac{3x^2/t^3-2\sqrt{3}\varrho\,x\,v/t^2+v^2/t}{2(1-\varrho^2)}\Big)\\ &=&\frac{\sqrt{3}}{\pi t^2}\,\exp\Big(-\frac{6x^2-6\,x\,v\,t+2v^2\,t^2}{t^3}\Big)\,. \end{eqnarray} This function satisfies the Kolmogorov PDE $$ \partial_t P=-v\,\partial_x P+\frac{1}{2}\partial_{vv}P $$ for the diffusion process $$ \left(\begin{array}{c}dx\\dv\end{array}\right)=\left(\begin{array}{c}v\\0\end{array}\right)\,dt+\left(\begin{array}{cc}0&0\\1&0\end{array}\right)\left(\begin{array}{c}dW\\dW_v\end{array}\right) $$ where $W_v$ is a dummy BM that is not driving anything.

The Book of Karatzas and Shreve (Brownian Motion and Stochastic Calculus) is very useful for such problems.

We know the covariance $E[\int_0^tW(u)\,du\int_0^tW(v)\,dv]=t^3/3\,.$ It is easy to see that the covariance between position $x(t)=\int_0^tW(u)\,du$ and velocity $v(t)=\dot x(t)=W(t)$ is $E[(\int_0^tW(u)\,du) W(t)]=t^2/2\,.$ Therefore, the correlation between $x$ and $v$ is

$$ \varrho=\frac{t^2/2}{\sqrt{t^3/3}\sqrt{t}}=\frac{\sqrt{3}}{2}\,. $$ Let's normalize the variables to make them standard normal: $$ \hat{x}(t)=\frac{x(t)}{\sqrt{t^3/3}}\,,~~\hat{v}(t)=\frac{v(t)}{\sqrt{t}}\,. $$ Then, $P(\hat x,\hat v,t)$ is the density of a bivariate normal distribution with correlation $\varrho\,$: $$ P(\hat x,\hat v,t)=\frac{1}{2\pi\sqrt{1-\varrho^2}}\,\exp\Big(-\frac{\hat x^2-2\varrho\,\hat x\,\hat v+\hat v^2}{2(1-\varrho^2)}\Big)\,. $$ Therefore, \begin{eqnarray}\label{ePDF} P(x,v,t)&=&\frac{\sqrt{3}}{2\pi t^2\sqrt{1-\varrho^2}}\,\exp\Big(-\frac{3x^2/t^3-2\sqrt{3}\varrho\,x\,v/t^2+v^2/t}{2(1-\varrho^2)}\Big)\\ &=&\frac{\sqrt{3}}{\pi t^2}\,\exp\Big(-\frac{6x^2-6\,x\,v\,t+2v^2\,t^2}{t^3}\Big)\,. \end{eqnarray}

We know the covariance $E[\int_0^tW(u)\,du\int_0^tW(v)\,dv]=t^3/3\,.$ It is easy to see that the covariance between position $x(t)=\int_0^tW(u)\,du$ and velocity $v(t)=\dot x(t)=W(t)$ is $E[(\int_0^tW(u)\,du) W(t)]=t^2/2\,.$ Therefore, the correlation between $x$ and $v$ is

$$ \varrho=\frac{t^2/2}{\sqrt{t^3/3}\sqrt{t}}=\frac{\sqrt{3}}{2}\,. $$ Let's normalize the variables to make them standard normal: $$ \hat{x}(t)=\frac{x(t)}{\sqrt{t^3/3}}\,,~~\hat{v}(t)=\frac{v(t)}{\sqrt{t}}\,. $$ Then, $P(\hat x,\hat v,t)$ is the density of a bivariate normal distribution with correlation $\varrho\,$: $$ P(\hat x,\hat v,t)=\frac{1}{2\pi\sqrt{1-\varrho^2}}\,\exp\Big(-\frac{\hat x^2-2\varrho\,\hat x\,\hat v+\hat v^2}{2(1-\varrho^2)}\Big)\,. $$ Therefore, \begin{eqnarray}\label{ePDF} P(x,v,t)&=&\frac{\sqrt{3}}{2\pi t^2\sqrt{1-\varrho^2}}\,\exp\Big(-\frac{3x^2/t^3-2\sqrt{3}\varrho\,x\,v/t^2+v^2/t}{2(1-\varrho^2)}\Big)\\ &=&\frac{\sqrt{3}}{\pi t^2}\,\exp\Big(-\frac{6x^2-6\,x\,v\,t+2v^2\,t^2}{t^3}\Big)\,. \end{eqnarray} This function satisfies the Kolmogorov PDE $$ \partial_t P=-v\,\partial_x P+\frac{1}{2}\partial_{vv}P $$ for the diffusion process $$ \left(\begin{array}{c}dx\\dv\end{array}\right)=\left(\begin{array}{c}v\\0\end{array}\right)\,dt+\left(\begin{array}{cc}0&0\\1&0\end{array}\right)\left(\begin{array}{c}dW\\dW_v\end{array}\right) $$ where $W_v$ is a dummy BM that is not driving anything.

We know the covariance $E[\int_0^tW(u)\,du\int_0^tW(v)\,dv]=t^3/3\,.$ It is easy to see that the covariance between position $x(t)=\int_0^tW(u)\,du$ and velocity $v(t)=\dot x(t)=W(t)$ is $E[(\int_0^tW(u)\,du) W(t)]=t^2/2\,.$ Therefore, the correlation between $x$ and $v$ is

$$ \varrho=\frac{t^2/2}{\sqrt{t^3/3}\sqrt{t}}=\frac{\sqrt{3}}{2}\,. $$ Let's normalize the variables to make them standard normal: $$ \hat{x}(t)=\frac{x(t)}{\sqrt{t^3/3}}\,,~~\hat{v}(t)=\frac{v(t)}{\sqrt{t}}\,. $$ Then, $P(\hat x,\hat v,t)$ is the density of a bivariate normal distribution with correlation $\varrho\,$: $$ P(\hat x,\hat v,t)=\frac{1}{2\pi\sqrt{1-\varrho^2}}\,\exp\Big(-\frac{\hat x^2-2\varrho\,\hat x\,\hat v+\hat v^2}{2(1-\varrho^2)}\Big)\,. $$ Therefore, \begin{eqnarray}\label{ePDF} P(x,v,t)&=&\frac{3}{2\pi t^2\sqrt{1-\varrho^2}}\,\exp\Big(-\frac{3x^2/t^3-2\sqrt{3}\varrho\,x\,v/t^2+v^2/t}{2(1-\varrho^2)}\Big)\\ &=&\frac{3}{\pi t^2}\,\exp\Big(-\frac{6x^2-6\,x\,v\,t+2v^2\,t^2}{t^3}\Big)\,. \end{eqnarray}

We know the covariance $E[\int_0^tW(u)\,du\int_0^tW(v)\,dv]=t^3/3\,.$ It is easy to see that the covariance between position $x(t)=\int_0^tW(u)\,du$ and velocity $v(t)=\dot x(t)=W(t)$ is $E[(\int_0^tW(u)\,du) W(t)]=t^2/2\,.$ Therefore, the correlation between $x$ and $v$ is

$$ \varrho=\frac{t^2/2}{\sqrt{t^3/3}\sqrt{t}}=\frac{\sqrt{3}}{2}\,. $$ Let's normalize the variables to make them standard normal: $$ \hat{x}(t)=\frac{x(t)}{\sqrt{t^3/3}}\,,~~\hat{v}(t)=\frac{v(t)}{\sqrt{t}}\,. $$ Then, $P(\hat x,\hat v,t)$ is the density of a bivariate normal distribution with correlation $\varrho\,$: $$ P(\hat x,\hat v,t)=\frac{1}{2\pi\sqrt{1-\varrho^2}}\,\exp\Big(-\frac{\hat x^2-2\varrho\,\hat x\,\hat v+\hat v^2}{2(1-\varrho^2)}\Big)\,. $$ Therefore, \begin{eqnarray}\label{ePDF} P(x,v,t)&=&\frac{\sqrt{3}}{2\pi t^2\sqrt{1-\varrho^2}}\,\exp\Big(-\frac{3x^2/t^3-2\sqrt{3}\varrho\,x\,v/t^2+v^2/t}{2(1-\varrho^2)}\Big)\\ &=&\frac{\sqrt{3}}{\pi t^2}\,\exp\Big(-\frac{6x^2-6\,x\,v\,t+2v^2\,t^2}{t^3}\Big)\,. \end{eqnarray}