An elevator is decending with acceleration 12m/s^2 what is d distance travelled by a body in lift in 2 second

Answer 1

dear, as acceleration of elevator is > 9.8 hence body will leave the surface of elevator. i.e. acceleration of lift is 12 while acceleration of body will 9.8 m/s^2. now distance covered by body will 1/2 *9.8*2^2.
thanks.

Answer 2

Since the elevator moves from one floor to the other, let us assume that it started from zero velocity. Now we know

S = Ut + 1/2 X (a X t^2)
Since U initial velocity = 0,
S = 1/2 X (a X t^2)
S = 1/2 X (12 X 2^2)
S = 24m

Hence the distance travelled by the body is 24m.

Answer 3

If the elevator started from rest
d=.5*a*t^2
in this case
d=24 m

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