Respuesta :
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
