Amodel rocket is fired vertically upward from rest. its acceleration for the first three seconds is a(t) = 72t, at which time the fuel is exhausted and it becomes a freely "falling" body. sixteen seconds later, the rocket's parachute opens, and the (downward) velocity slows linearly to −28 ft/s in 5 s. the rocket then "floats" to the ground at that rate. (a) determine the position function s and the velocity function v (for all times t

Velocity function = V(T)= Integral of 72t

= ∫72xdx

= 72∫xdx

Now solving for ∫xdx

Apply the power rule

∫x^ndx = xn + 1/ n + 1 with n=1:

= x^2 / 2

Plug that in the solved integrals:

= 72∫xdx

= 36 x^2

So…

72t^2/2= 36(t^2) = this is the velocity function

Position = S(T)

= Integral of 36(t^2)

= ∫36x^2dx

= 36∫x^2dx

Now solving for ∫x^2dx

Apply the power rule

∫x^ndx = xn + 1/ n + 1 with n = 2:

= x^3 / 3

Plug that in the solved integrals:

= 36∫x^2dx

= 12x^3

So the position of S(t) = 12 (t^3)

answer; i believe the correct answer is (40,000 j) good luck