The slevator starts at 1st floor and accelerated 1m/s^2 for 6 seconds and continues at a constant velocity for 12 seconds more and is then stopped in 4 seconds with constant deceleration. If the floors are 3 meters apart at what floor does the elevator stops?

Answer 1

with what constant deacceleration value?

Answer 2

Ok, I see it now, the total displacement is the sum of

d1=1/2 *1*6^2
d2= 1*6*12
d3=1/2 (-1*6*12)/4 *4^2

check those...

then floor=displacement/3 + 1

assuming ground floor is called floor 1

displacement= d1+d2+d3

Answer 3

Hmmm. I get
d3 = 6*4 + 1/2 (-6/4)*4^2
since v goes from 6 to 0 in 4 seconds.

bob's the physicist, but ...