1.

In the figure, the pulleyP moves to the rightwitha constant speed u. The downward spee of A isv_(A) andthe speedof Bto therightisv_(B) . Then,

Answer»

`v_(A)= v_(B)`
`v_(B)= u+v_(A)`
`v_(B) + u =v_(A)`
the two blocks have accelerationsof the same magnitude.

Solution :At anyinstant of time,let thelengthof the STRING `BP=l_(1)` and the LENGTH ` PA= l_(2)`. In a furthertime t, let B moveto the rightby x and A move down by y,while P moves to the rightby ut. As the lengthof the string must remain constant ,
` l_(1) + l_(2) = ( l_(1) -x + ut) + (l_(2) +y ) `
` orx = ut+ yor (dx)/(dt) = u + (dy)/(dt) `
` (dx)/(dt) ` = SPEED of B to the right `= v_(B)`
` (dy)/(dt) ` = downwardspeed of A` = v_(A) therefore v_(B) = u+V_(A)`
Also `d_(v_(B))/(dt) = (dv_(A))/(dt) = or a_(B) = a_(A)`


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