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?

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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