Moving Dots

Let $x$ and $t$ be the points and $D = |x-t|.$ Then the variance is $E(D^2)-E(D)^2 = \frac19 \int_0^3 \int_0^3 (x-t)^2 \, dx \, dt - \left( \frac19 \int_0^3 \int_0^3 |x-t| \, dx \, dt \right)^2.$ It's well-known that $E(D) = 1$ by a symmetry argument, but it's not hard to derive it from this integral: $\int_0^3 |x-t| \, dx$ is the area of two isoceles right triangles (draw the picture): $t^2/2 + (3-t)^2/2.$ The integral of that over $[0,3]$ is $27/6 - (-27/6) = 9,$ and multiplying by $1/9$ gives $E(D) = 1.$

As for $E(D^2),$ this is a similarly straightforward computation, which I will leave to the reader: $E(D^2) = 3/2.$ So the answer is $E(D^2)-E(D)^2 = 1/2,$ so the correct choice is $\fbox{50}.$