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

The question asks for the determination of a model rocket's position and velocity functions over time, accounting for powered flight, free-fall, and parachute-assisted descent.

Step-by-step explanation:

The student has described a model rocket problem which involves calculation of the rocket's position and velocity as functions of time in different phases of its flight. The rocket's motion is divided into three segments: first, the powered ascent with a given acceleration function a(t) = 72t, second, the free-fall period, and lastly, deceleration and constant descent due to the parachute deployment. To solve for the position function s(t) and the velocity function v(t), knowledge of kinematic equations and integration of acceleration to get velocity, and again to get position is essential.

Answer 2

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)