Question

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.

Answer 1

The shortest time to lift the car to the fourth floor of the parking garage, starting from rest and ending at rest, is 46 seconds.

Step-by-step explanation:

To determine the shortest time to lift the car to the fourth floor of the parking garage, we can break down the problem into different stages: acceleration, constant velocity, and deceleration. First, let's calculate the time taken to accelerate from rest to the maximum speed. The elevator can accelerate at a rate of 0.6 ft/s^2, and the maximum speed is 8 ft/s. Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time:

v = u + at

8 = 0 + 0.6t

t = 8/0.6 = 13.33 seconds

Next, let's calculate the time taken to decelerate from the maximum speed to rest. The elevator decelerates at a rate of 0.3 ft/s^2. Using the same equation:

v = u + at

0 = 8 + (-0.3)t

t = -8/-0.3 = 26.67 seconds

Finally, we need to calculate the time taken to travel the remaining distance at a constant speed. The distance to the fourth floor is 48 ft, and the elevator is already at the maximum speed of 8 ft/s. Using the equation d = vt, where d is the distance, v is the velocity, and t is the time:

t = d/v

t = 48/8 = 6 seconds

Therefore, the shortest time to make the lift, starting from rest and ending at rest, is 13.33 + 6 + 26.67 = 46 seconds.

Answer 2

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