Answer:
a) Δx = 49.23 mi
, b) Δx = 5.77 mi
Explanation:
As we have an acceleration function we must use the definition of kinematics
a = dv / dt
∫dv = ∫ a dt
we integrate and evaluators
v - vo = ∫ (-1280 (1 + 8t)⁻³ dt = -1280 ∫ (1+ 8t)⁻³ dt
We change variables
1+ 8t = u
8 dt = du
v - v₀ = -1280 ∫ u⁻³ du / 8
v -v₀ = -1280 / 8 (-u⁻²/2)
v - v₀ = 80 (1+ 8t)⁻²
We evaluate between the initial t = 0 v₀ = 80 and the final instant t and v
v- 80 = 80 [(1 + 8t)⁻² - 1]
v = 80 (1+ 8t)⁻²
We repeat the process for defining speed is
v = dx / dt
dx = vdt
x-x₀ = 80 ∫ (1-8t)⁻² dt
x-x₀ = 80 ∫ u⁻² dt / 8
x-x₀ = 80 (-1 / u)
x-x₀ = -80 (1 / (1 + 8t))
We evaluate for t = 0 and x₀ and the upper point t and x
x -x₀ = -80 [1 / (1 + 8t) - 1]
We already have the function of time displacement
a) let's calculate the position at the two points and be
t = 0 h
x = x₀
t = 0.2 h
x-x₀ = -80 [1 / (1 +8 02) -1]
x-x₀ = 49.23
displacement is
Δx = x (0.2) - x (0)
Δx = 49.23 mi
b) in the interval t = 0.2 h at t = 0.4 h
t = 0.4h
x- x₀ = -80 [1 / (1+ 8 0.4) -1]
x-x₀ = 55 mi
Δx = x (0.4) - x (0.2)
Δx = 55 - 49.23
Δx = 5.77 mi
