Sliding linked blocks

$(A_x - B_x)^2 + (A_y - B_y)^2 = L_{AB}^2$

Time-differentiating both sides:

$(A_x - B_x)(\dot{A}_x - \dot{B}_x) + (A_y - B_y)(\dot{A}_y - \dot{B}_y) = 0 \\ (A_x - B_x)(0 - \dot{B}_x) + (A_y - B_y)(\dot{A}_y - 0) = 0$

It is easy to calculate $(A_x, B_x, A_y, B_y)$ with respect to point $D$ using the information given. All that remains is to re-arrange and solve for $\dot{B}_x$ , which turns out to be $3$

Let the length of the rod AB be $l$ and the angle it subtends with horizontal be $α$ . Then $V_a=lcosα\dfrac{dα}{dt}=4$ . This implies $l\dfrac{dα}{dt}=5$ (since $cosα=\dfrac{4}{5}$ ). Then $|V_b|=lsinα\dfrac{dα}{dt}=3$ (since $sinα=\dfrac{3}{5}$ )