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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.

Respuesta :

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

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