The machine as shown has 2 rods of length 1m connected by a pivot...

${v}_{A} = \dfrac{d x}{d t} = \dfrac{d \left[2lcos\left(\theta\right)\right]}{d t} = - 2lsin\left(\theta\right)\dfrac{d \theta}{d t}$ and

Vertical velocity of $B$, {v}_{B} = \dfrac{d y}{d t} = \dfrac{d \left[lsin\left(\theta\right)\right]}{d t} = lcos\left(\theta\right)\dfrac{d \theta}{d t}$</p><p>$\dfrac{{v}_{A}}{{v}_{B}} = - 2\tan \left(\theta\right) ⇒ {v}_{B} = - \dfrac{{v}_{A}\cot \left(\theta\right)}{2}$<br /> <br />$$\begin{array}{ll}\theta \to\end{array}$$increase</p><p>$cot\theta \to$decrease</p><p>Hence, \left|{v}_{B}\right| \to$decrease.