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.) 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

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