In the figure shown,`a_(3) = 6 m//s^(2)` (downwards) and `a_(2) = 4 m//s^(2)` (upwards) .Find acceleration of `1`

Options

A. `1m//sec^(2)upwards`
B. `2m//sec^(2)upwards`
C. `1m//sce^(2)downwards`
D. `2//sec^(2)downwards`

Answer

A

Explanation

Since the points `1,2,3` and are movable so let their displacements are `x_(1),x_(2),x_(3)` and `x_(4)` We observe that length of the string between 1 and 4 and 2 and 3 are constants
`x_(1) +x_(4) = l_(1), (x_(2) -x_(4)) + (x_(3)-x_(4)) =l_(2)`
Differentiating twice w.r.t time we get
`a_(1) +a_(4) =0 rArr a_(1) = - a_(4) rarr (1)`
`a_(2) -a_(4) + a_(3) -a_(4) =0`
`rArr a_(2) +a_(3) -2a_(4) =0rarr(2)`
`a_(2) =-4ms^(-2) ,a_(3) = 6ms^(-2) , -4 + 6-2a_(4) =0`
`2a_(4) = 2rArr a_(4) = 1ms^(-2),` From (1)
`a_(1) = - a_(4) , a_(1) = - 1ms^(-2) ,a_(1) = 1ms^(-2)` upwards .