A block is kept the floor of an elevator at rest. The elevator starts descending with an acceleration of 12m//s^(2). Find the displacement of the block during the first 1s after the start.

A block is kept the floor of an elevator at rest. The elevator starts

1.	A block is kept the floor of an elevator at rest. The elevator starts descending with an acceleration of 12m//s^(2). Find the displacement of the block during the first 1s after the start.
Answer» Solution : `mg-N=ma_(0)` `N=m(g-a_(0))` Here `a_(0)gtg`, the block will not be in contact with the FLOOR For the block`:` `X=(1)/(2)g t^(2)=(1)/(2)xx10xx1^(2)=5m` In this duration, the elevator will fall a DISTANCE `s'=(1)/(2)a_(0)t^(2)=(1)/(2)xx12xxt^(2)=6m`