The position of an object as a function of time is given by [x (t) = at^3 - bt^2 + ct - d] where (a = 3.6, b = 4.0, c = 60 m, and d = 7.0). (a) Find the instantaneous acceleration at t = 2.4 s. (b) Find the average acceleration over the first 2.4

(a) a = 43.84 m/s²

(b) am = 17.92 m/s²

Explanation:

x (t) = at³ - bt² + ct - d Equation (1)

We replace a = 3.6, b = 4.0, c = 60 m, and d = 7.0 in the Equation (1)

x (t) = [ (3.6) t³ - (4) t² + (60) t - 70]m : Position of the object as a function of time

v (t) = dx/dt = [ (10.8) t² - 8t + 60] (m/s) : speed of the object as a function of time

a (t) = dv/dt = [ (21.6) t - 8] (m/s²) : acceleration of the object as a function of time

(a) Instantaneous acceleration at t = 2.4 s

We replace t = 2.4 s in the equation of the acceleration:

a (t) = (21.6) t - 8

a (t=2.4s) = (21.6) (2.4) - 8

a (t=2.4s) = 43.84 m/s²

(b) Average acceleration over the first 2.4 s

We known v (t) = [ (10.8) t² - 8t + 60] (m/s)

am: average acceleration

am = Δv/Δt = (v₂-v₁) / (t₂-t₁)

v₂ = v (t=2.4s) = [ (10.8) (2.4) ² - 8 (2.4) + 60] = 62.208 - 19.2 + 60 = 103.008 m/s

v₁ = v (t=0s) = [ (10.8) (0) ² - 8 (0) + 60] = 60 m/s

t₂ = 2.4 s

t₁ = 0 s

am = Δv/Δt = (103.008 - 60) / (2.4 - 0)

am = 17.92 m/s²