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5 February, 17:11

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.

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  1. 5 February, 18:48
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    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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