At a certain time velocities of `1` and `2` both are `1m//s` 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_(4)=l_(2)` ( length of second string)
`rArrv_(1)+v_(4)=0& v_(2)+v_(3)-2v_(4)=0`
`rArrv_(2)+v_(3)+2v_(1)=0`
Taking upward direction as positive `v_(1)=v_(2)=1`
so `1+v_(3)+2xl=0rArrv_(3)=-3 ms^(-1)`
i.e. velocity of block 3 is `3ms^(-1)` downwards.