Answer to Question #303019 in Classical Mechanics for Elevator

Question

The elevator in an office building, starting from rest at the first floor, is accelerated 0.75 m/sec2 for 5 secs. It continues at constant velocity for 12 secs more and is then stopped in 3 secs with constant deceleration. If the floors are 3.75m apart, at what floor did the elevator stop?

Answer 1

$n=\frac h{3.75}=\frac 1{3.75}(\frac{v^2}{2a}+\frac{at_1^2}2+vt_2)=14.$