Question

A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds is a(t) = 96t, at which time the fuel is exhausted and it becomes a freely "falling" body. Nineteen seconds later, the rocket's parachute opens, and the (downward) velocity slows linearly to −16 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).
b) At what time does the rocket reach its maximum height?

What is that height? (Round your answer to the nearest integer.)

c) At what time does the rocket land? (Round your answer to one decimal place.)

Answer 1

Answer:

In kinematics questions we need to separate the question into different parts if the acceleration changes. Here, there are three time intervals where acceleration is different.

1) a(t) = 96t. We can find the velocity function of the rocket by integrating the acceleration function. Then we can integrate again to find the position function.

$v(t) = \int{a(t)} \, t = \int {96t} \, dt = 48t^2 + C$

'C' is the integration constant. We can find this constant by investigating the initial conditions.

$v(t = 0) = 48(0)^2 + C = 0\\C = 0$

We know that the rocket is initially at rest, so 'C' should be zero.

$s(t) = \int {v(t)} \, dt  = \int {48t^2} \, dt = 16t^3 + C$

Again, the rocket started from ground zero, so C = 0.

We should conclude the first part by calculating the final position and final velocity of the rocket.

$s(t=3) = 16(3)^3 = 432ft\\v(t=3) = 48(3)^2 = 432ft/s$

2) For the second part, the rocket is in free fall, so

$a(t) = -32.2ft/s^2\\v(t) = -32.2t + C\\v(t=3) = -32.2*3 + C = 432\\C = 335.4\\v(t) = -32.2t + 335.4\\s(t) = -16.1t^2 + 335.4t + C\\s(t=3) = -16.1(3)^2 + 335.4*3 + C = 432\\C = 432 + 144.9 - 1006.2 = -429.3\\s(t) = -16.1t^2 + 335.4t - 429.3$

The maximum height that the rocket reaches is when its velocity is zero.

So,

$v(t) = -32.2t + 335.4 = 0\\t = 10.4 s$

The maximum height is

$s(t=10.4) = -16.1t^2 + 335.4t - 429.3 = -16.1(10.4)^2 + 335.4*10.4 - 429.3 = -1741.3 + 3488.1 - 429.3 = 1318 ~ft$

The final positions for the part 2 is

$s(t=19) = -16.1(19)^2 + 335.4*19 - 429.3 = -5812.1 + 6372 - 429.3 = 131.2~ft\\v(t=19) = -32.2*19 + 335..4 = -276.4~ft/s$

3) With the parachute, the velocity is dropped from -276.4 to 16 in 5 s.

$a(t) = \frac{-16 - (-276.4))}{5} = 52ft/s^2\\v(t) = 52t + C\\v(t= 19) = 52*19 + C= -276.4\\988+ C = -276.4\\C = -1264.4\\v(t) = 52t - 1264.4$

$s(t) = 26t^2 - 1264.4t + C \\s(t=19) = 26(19)^2 - 1264.4*19 + C = 131.2\\9386 - 24023.6 + C = 131.2\\C = 14768.8\\s(t) = 26t^2 - 1264.4t + 14768.8$

The rocket lands

$s(t) = 26t^2 - 1264.4t + 14768.8 = 0\\t = 29s.$