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)

6 cos 34, 6 sin 34 trig

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this