A rod of length 'l' is inclined at an angle 'theta' with the floor against a smooth vertical wall. If the end A moves instantaneously with velocity `v_(1)` what is the velocity of end `B` at the instant when rod makes 'theta' angle with the horizontal .

Explanation

Let at any instant, end `B` and `A` are at a distance x and y respectively from the point 'O'
Thus we have `x^(2) +y^(2) = l^(2)…..(1)`
Here l is the length of the rod which is constant
Differentiating eq (1) with respect to time, we get
`(d)/(dt)(x^(2)+y^(2))= (d)/(dt)(l^(2)), 2x (dx)/(dt) + 2y(dy)/(dt)=0`
If `(dx)/(dt) =v_(2)` and `(dy)/(dt) =-v_(1)`
`x(v_(2)) +y(-v_(1)) =0`
`rArrv_(2)=((y)/(x))v_(1)=v_(1)tantheta`
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