Suppose $\{W^{(1)}_t, t\geq 0\}$ and {$W^{(2)}_t, t\geq 0\}$ are two independent Brownian motions. If I recall correctly, the distance between the two at a given time has the following property:
$$\mathbb{E}\big|W^{(1)}_s-W^{(2)}_s\big|=\sqrt{s}$$
I'm having a hard time finding this in a book, but I'm pretty sure it's right. Can you explain why or tell me why I'm wrong?
Consider the process $$X_s = W^{(1)}_s-W^{(2)}_s$$
At any given time, this is the difference of two mean zero, independent, normals with variance $s$. That means $X_s$ is a mean zero normal with variance $2s$.
So the question is, what is the expected value of the absolute value of a normal random variable. This is called a Folded Normal Distribution. One that was folded from a normal with mean $0$ and variance $\sigma$ has mean $$\sigma \sqrt{\frac{2}{\pi}}$$ So in your case, $$\sqrt{2s}\sqrt{\frac{2}{\pi}} = 2 \sqrt{\frac{s}{\pi}}$$