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 u. As a result of this, end A starts moving down along the wall. Find the velocity of the other end B downward when the rod makes an angle `theta` with the horizontal.

Explanation

Let us first find the relation between the two displacements. Then differentiate with respect to time. Here if the distance from the corner to the point A is x and that up to B is y, the left velocity of pointA can be given as `v_(A)=(dx)//(dt)` and that of B can be given as `v_(B)=-(dy)//(dt)` (negative sign indicates y decreasing).
If we relate x and y: `x^(2)+y^(2)=l^(2)`
Differentiating with respect to t, `2x(dx)/(dt)+2y(dy)/(dt)=0`
`implies xv_(A) = yv_(B) =xu=yv_(B) implies v_(B) = u(x)/(y)=u cot theta`
Alternatively: In cases where distance between two points is always fixed, we can say the relative velocity of one point of an object with respect to any other point of the same object in the direction of the lime joining them will always remain zero, as their separation always remains constant.
Here, in the above example, the distance between the points A and B of the rod always remains constant, thus, the two points must have the same velocity components in the direction of their line joining, i.e., along the length of the rod.
If point B is moving down with velocity `v_(B)` its component along the length of the rod is `v_(B) sin theta`. Similarly the voelcity components of point A along the length of rod is `u cos theta`. Thus, we have `v_(B) sin theta=u cos theta` or `v_(B)=u cot theta`.