Figure shows a rod of length $l$ resting on a wall and the floor. Its lower end A is pulled towards left with a constant velocity $v$. Find the velocity of the other end B downward when the rod makes an angle $\theta$ with the horizontal.

Solution In such type of problems, when velocity of one part of a body is given and that of other is required, we first find the relation between the two displacements, then differentiate them with respect to time. Here, if the distance from the corner to the point $A is x$ and that up to $B$ is $y$. Then,

$u=\frac{dx}{dt}$ and

$v_B=-\frac{dy}{dt}$

Further, $x^2+y^2=l^2$

Differentiating with respect to time $t.$

$$\begin{gathered} 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0 \\ xv=y{v}_B \\ {v}_B=\frac{x}{y}u=ucot\theta \end{gathered}$$