Question

A model 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

Answer 1

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)