A car is to be hoisted by an elevator to the fourth floor of a parking garage, which is 48 ft above the ground. if the elevator can accelerate at 0.6 ft/s 2, decelerate at 0.3 ft/s 2, and reach a maximum speed of 8 ft/s, determine the shortest time to make the lift, starting from rest and ending at rest.

Create a velocity - time diagram as shown below.

The car accelerates from rest to a velocity, v, followed by deceleration to rest at the expected height.

Acceleration phase:

The velocity v is attained in time t₁ with acceleration of 0.6 ft/s², therefore

v = (0.6 ft/s²) (t₁ s) = 0.6t₁ ft/s

t₁ = v/0.6 = 1.6667v s

The distance traveled is

h₁ = (1/2) * (0.6 ft/s²) * (t₁ s) ²

= 0.3 (1.6667v) ²

= 0.8334v² ft

Deceleration phase:

The car comes to rest from an initial velocity of v with a deceleration of 0.3 ft/s² in time t₂. Therefore

v - (0.3 ft/s²) * (t₂ s) = 0

t₂ = v/0.3 = 3.3333v s

The distance traveled is

h₂ = vt₂ - (1/2) * (0.3 ft/s²) * (t₂ s) ²

= 3.3333v² - 0.15 * (3.3333v) ²

= 1.6667v² ft

The total distance traveled should be 48 ft, therefore

0.8334v² + 1.6667v² = 48

2.5v² = 48

v = 4.3818 ft/s (should not exceed 8 ft/s, so it is okay).

The shortest time to make the lift is

t₁ + t₂ = 1.6667v + 3.3333v

= 5v

= 5*4.3818

= 21.9 s

Answer: 21.9 s