An atwood machine is setup in an elevator moving upward at `5 m//s` and slowing down at `2m//s^(2)` The initial velocity of block B is `2m//s` upward and the acceleration of block A is `3 m//s^(2)` downwards Find the time (in sec) at which block B will return to its initial position. Assume the string remains taut and the acceleration of the elevator does not change during the required time interval

Answer

`4 sec`

Explanation

`v_(B)+v_(A)=2v_(P)=2xx5`
`2+v_(A)=10 Rightarrow v_(A)=8 uparrow`
`a_(B)+a_(A)=2a_(P)=2xx-2`
`a_(B) -3=-4`
`a_(B)=-1`
`s=0=2xxt-1/2xx1xxt^(2)`
`t=4 sec`