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8 December, 19:17

Two elevators begin descending from the same height. Elevator A has descended 4 feet after one second, 9 feet after two seconds, 14 feet after three seconds, and so on. Elevator B has descended 3.5 feet after one second, 6.5 feet after two seconds, 9.5 feet after three seconds, and so on. How many feet would each elevator descend in 10 seconds?

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Answers (2)
  1. 8 December, 22:34
    0
    A: 49 ft; B: 30.5 ft

    Explanation:

    correct answer
  2. 8 December, 22:57
    0
    a) y = 400 ft b) y = 350 ft

    Explanation:

    We can solve that problem with kinematic relationships

    y = v₀ t + ½ a t²

    Let's apply this equation to the elevator 1

    y₁ = v₀ t₁ + ½ to t₁²

    time t = 1 s

    y₁ = v₀ 1 + ½ a 1²

    y₁ = v₀ + ½ a

    For t = 2s

    y₂ = v₀ 2 + ½ a 2²

    y₂ = 2 v₀ + 2 a

    Let's write the equations and solve the system

    4 = v₀ + ½ a

    9 = 2 v₀ + 2 a

    Let's multiply the first by - 2

    -8 = - 2v₀ - a

    9 = 2v₀ + 2 a

    Let's add

    1 = a

    We replace in the first

    4 = v₀ + ½ 1

    v₀ = 4 - 1/2

    v₀ = 3.5

    The equation for the first elevator is

    y = 3.5 t + ½ t²

    For t = 10 s

    y = 3.5 10 + ½ 10²

    y = 400 ft

    We repeat the process for the second elevator

    t = 1s

    y₁ = v₀ 1 + ½ a 1²

    3.5 = v₀ + ½ a

    t = 2 s

    y₂ = v₀ 2 + ½ a 2²

    6.5 = 2 v₀ + 2 a

    multiply by - 2

    -7 = - 2 v₀ - a

    6.5 = 2 v₀ + 2 a

    Let's add

    -0.5 = a

    I replace in the first equation

    3.5 = v₀ + ½ (-0.5)

    v₀ = 3.5 + 0.25

    v₀ = 3.75

    The equation is

    y = 3.75 t - 0.25 t²

    For t = 10s

    y = 3.75 10 - 0.25 10²

    y = 350 ft
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