A rod $AB$ of length $3m$ which remains in the same vertical plane has its ends $A$and $B$ constrained to remain in contact with a horizontal floor and a vertical wall respectively. Determine the velocity of the end B at the position, if the point $A$has a velocity of $2 m/s$ and an acceleration of $1.6 m/s^2$ rightward.

The velocity of end A :

The motion of $B$ can be looked as; translation of end $B$ with velocity $v_A$ and rotation of $B$ about  A with tangential velocity $v_{BA}$. If $\omega$ be the angular velocity of $B$ about$A$, then $v_{BA}=\omega \times 3=3\omega$ . Since the velocity vector of end $B$ moves downward, so resultant of ${\hat{v}}_A$ and ${\hat{v}}_{BA}$ must be equal to ${\hat{v}}_B$ (see figure).

From figure $\left(b\right)$, $\frac{v_A}{v_B}=\tan 60°$

$$\begin{gathered} \therefore v_B=\frac{v_A}{\tan 60°}=\frac{2}{\tan 60°} \\ =1.16 m/s \end{gathered}$$