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An elevator car whose floor to ceiling d...

An elevator car whose floor to ceiling distance is equal to `2.7m` starts ascending with constant acceleration `1.2 m//s^2.` 2 s after the start, a bolt begins falling from the ceiling of the car. Find
(a)the time after which bolt hits the floor of the elevator.
(b)the net displacement and distance travelled by the bolt, with respect to earth. (Take `g=9.8 m//s^2)`

Text Solution

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(a) If we consider elevator at rest, then acceleration of the bolt with respect to elevator is,
`a_(BE) = - (9.8 + 1.2) = - 11 m//s^(2)` (downwards)
After 2s velocity of lift is `v = at 1.2 xx 2 = 2.4 m//s`
Therefore, initial velocity of the bolt is also 2.4 m/s and it gets accelerated with relative acceleration `11 m//s^(2)` with respect to elevator initial velocity of bolt is zero and it has to travel `2.7 m` with `11 m//s^(2)`. Thus, time taken can be directly given form equation, `s_(BE) = (1)/(2) a_(BE) t^(2)`
`t = sqrt((2s_(BE))/(a_(BE)) ) = sqrt((2 xx 2.7)/(11)) = 0.7 s`
(b) Displacement of bolt relative to ground in 0.7 s
`s = u_(B) t + (1)/(2) a_(B) t^(2)``= 2.4 xx 0.7 + (1)/(2) (-9.8) xx (0.7)^(2)`
`s = - 0.72 m`

Velocity of bolt will become zero after a time
`t_(0) = (u_(B))/(g) = (2.4)/(9.8) = 0.245 s`
Therefore, distace travelled by the bolt `=s_(1) + s_(2)`
`= (u_(B)^(2))/(2g) + (1)/(2) g (t - t_(0))^(2)`
`= ((2.4)^(2))/(2 xx 9.8) + (1)/(2) xx 9.8 xx (0.7 - 0.245)^(2) = 1.3 m`
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