Answer:
Explanation:
From the given information:
Let's assume that the missing function is:
s(t) = t³ - 6t², t ≥ 0
From part (b), we are to find the given  required terms when time t = 2
So; from the function s(t) =  t³ - 6t², t ≥ 0
[tex]velocity \ v(t) \ = \dfrac{d}{dt}s(t)[/tex]
[tex]velocity \ v(t) \ = \dfrac{d}{dt}(t^3 - 6t^2)[/tex]
[tex]velocity \ v(t) \ = 3t^2 - 12t[/tex]
[tex]acceleration a(t) = \dfrac{d}{dt}*v(t)[/tex]
[tex]acceleration a(t) = \dfrac{d}{dt}(3t^2 - 12 t)[/tex]
[tex]acceleration\ a(t) = 6t - 12[/tex]
At time t = 2
The position; S(2) = (2)² - 6(2)²
S(2) = 8 - 6(4)
S(2) = 8 - 24
S(2) = - 16 ft
v(2) = 3(2)² - 12 (2)
v(2) = 3(4) - 24
v(2) = 12 - 24
v(2) = - 12 ft/s
speed = |v(2)|
|v(2)| Â = |(-12)|
|v(2)| = 12 ft/s
acceleration = 6t - 12
acceleration = 6(2) - 12
acceleration = Â 12 - 12
acceleration =  0 ft/s²
