BTW ... this problem ain't calculus, it's just plain old mechanics using Newton's 2nd law and kinematics.
using g = 9.8 m/s<sup>2</sup> ...
tension in the cable, T = 25.2 kN
weight of the elevator, W = (1500 kg)(9.8 m/s<sup>2</sup>) = 14.7 kN
F<sub>net</sub> = ma
accelerating upward, tension > weight ...
T - W = ma
(T - W)/m = a
(25200 - 14700)/1500 = 7 m/s<sup>2</sup>
accelerating to 5 m/s from rest at 7 m/s<sup>2</sup> will take 5/7 sec
total displacement = 25m
total time = 6 sec
let t = time for deceleration
6 - t - (5/7) = (35/7) - t = time traveling at a constant speed
total distance = distance accelerating + distance at a constant speed + distance decelerating
using the kinematics equations d = (1/2)at<sup>2</sup>, d = vt, and
d = (1/2)(v<sub>0</sub> + v<sub>f</sub>)t ...
25 = (1/2)(7)(5/7)<sup>2</sup> + 5[(35/7) - t] + (1/2)(5 + 0)t
solving for t ...
25 = 25/14 + 175/7 - 5t + (5/2)t
25 = 375/14 - (5/2)t
350 = 375 - 35t
t = 5/7 sec
so ... the magnitude of deceleration will match the magnitude of acceleration (7 m/<sup>2</sup>)
F<sub>net</sub> = ma
during deceleration, weight > tension ...
W - T = ma
W - ma = T
14700 - (1500)(7) = T
4200 N = T
4.2 kN = T
