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: 49 ft; B: 30.5 ft

Explanation:

correct answer

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