At certain moment of time, velocities of 1 and 2 both are `1 ms^(-1)` upwards. Find the velocity of 3 at that moment.

Explanation

`x_(1)+x_(4)=l_(1)` (length of first string)
`x_(2)-x_(4)+x_(3)-x_(3)-x_(4)=l_(2)` (length of second string)
`impliesv_(1)+v_(4)=0&v_(2)+v_(3)-2V_(4)=0`
`impliesv_(2)+v_(3)+2v_(1)=0`
Thinking upward direction s positive
`v_(1)=v_(2)=1`
`so 1+v_(3)+2xx1=0impliesv_(3)= -3 ms^(-1)`
i.e., velocity of block 3 is `3 ms^(-1)` downwards.