Problem 1: a car is to be hoisted by elevator to the fourth floor of a parking garage, which is 52 ft above the ground. if the elevator can accelerate at 0.7 ft/s2, decelerate at 0.35 ft/s2, and reach a maximum speed of 8 ft/s, determine the shortest time to make the lift, starting from rest and ending at rest.

The working equation for this problem is written below:
x = v₀t + 0.5at², where x is the distance traveled, v₀ is the initial velocity, a is the acceleration and t is the time

Let's apply the concept of calculus. The maximum speed is equated to the derivative of x with respect to t.

dx/dt = 8 ft/s = v₀ + at

Since the it starts from rest, v₀ = 0

8 = at
t = 8/a  

The net acceleration is 0.7 - 0.35 = 0.35 ft/s². Thus,

t = 8/0.35 = 22.86 seconds