Question

A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds is a(t)=60t, at which time the fuel is exhausted and it becomes a freely "falling" body. Fourteen seconds later, the rocket's parachute opens, and the (downward) velocity slows linearly to -18ft/sec in 5 seconds. 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'). Sketch the graphs of 's' and 'v'. (b) At what time does the rocket reach its maximum height, and what is that height? (c) At what time does the rocket land? Please use antiderivatives to solve the problem.

Answer 1

Answer:

See steps below

Step-by-step explanation:

We have

a(t) = 60t for 0 ≤ t ≤ 3.

V(t) is the anti-derivative of a(t), so

$\large v(t)=\frac{60t^2}{2}+C=30t^2+C$

where C is a constant.

But v(0) = 0 (the rocket is launched from rest), so C = 0 and

$\large v(t)=30t^2\;(0\leq t<3)$

S(t) is the anti-derivative of v(t), so

$\large s(t)=\frac{30t^3}{3}+C=10t^3+C\;(0\leq t\leq 3)$

where C is a constant.

But s(0) = 0 (the rocket is on the land), so C = 0 and

$\large s(t)=10t^3\;(0\leq t\leq 3)$

After 3 seconds the fuel is exhausted and it becomes a freely "falling" body for 14 seconds until the parachute opens.

Hence from t=3 until t=17 the acceleration is the gravity. So

$\large a(t)=-32.174\;ft/sec^2$

the anti-derivative v(t) is now

v(t) = -32.174t + C

where C is a constant.

But v(3) = 270 ft/sec, in consequence:  

270 = -32.174(3) +C and C = 366.522 and

v(t) = -32.174t + 366.522 ( 3 ≤ t ≤ 17)

The anti-derivative of v(t) is now

$\large s(t)=-32.174\frac{t^2}{2}+366.522t+C=-16.087t^2+366.522t+C$

But

$\large  s(3)=10(3)^3=270$

Hence

$\large 270=-16.087(3)^2+366.522*3+C\Rightarrow C=-684.783$

and

$\large s(t)=-16.087t^2+366.522*t-684.783$ for  3≤ t ≤ 17

At second 17 the parachute opens and the rocket gets the acceleration of  

$\large a(t)=-\frac{18}{5}=-3.6\;ft/sec^2$

until it lands.

In this case the anti-derivative is

v(t) = -3.6t + C for t  ≥  17

But  

v(17) = -32.174*17 + 366.522 = -180.436

so -3.6(17)+C = -180.436

hence C = -119.236  and

v(t) = -3.6t -119.236 (t  ≥  17 until landing)

for this v(t) the anti-derivative is

$\large s(t)=-3.6\frac{t^2}{2}-119.236t+C=-1.8t^2-119.236t+C$

since  

$\large s(17) = -16.087(17)^2+366.522*17-684.783=896.948$

then

$\large 896.948=-1.8(17)^2-119.236*17+C\Rightarrow C=3444.16$

 

and

$\large s(t)=-1.8t^2-119.236t+3444.16\;(t\geq 17)$

until landing.

When the rocket lands, s(t) becomes zero. It happens at the positive value of t such that

$\large -1.8t^2-119.236t+3444.16=0$

solving the quadratic equation we get t = 21.7463 seconds.  

So

$\large s(t)=-1.8t^2-119.236t+3444.16$ for 17 ≤ t ≤ 21.7463

Summarizing

$\large v(t)=\begin{cases}30t^2 & 0\leq t \leq 3\\-32.174t + 366.522 & 3\leq t \leq 17\\-3.6t -119.236&17\leq t \leq 21.7463 \end{cases}$

and

$\large s(t)=\begin{cases}10t^3 & 0\leq t \leq 3\\-16.087t^2+366.522*t-684.783 & 3\leq t \leq 17\\-1.8t^2-119.236t+3444.16&17\leq t \leq 21.7463\end{cases}$

The graphs of v(t) and s(t) are sketched in the pictures attached

(see pictures)

(b) At what time does the rocket reach its maximum height, and what is that height?

The rocket reaches its maximum height when v(t) = 0.

That happens at an instant t between 3 and 14. That is to say, at the instant t such that

v(t) = -32.174t + 366.522 =0

and t = 366.522/32.174 = 11.39187 seconds

The maximum height would be then s(11.39187)

$\large s(11.39187)=-16.087(11.39187)^2+366.522*11.39187-684.783=1402.902\;ft$

(c) At what time does the rocket land?

When the rocket lands s(t) becomes zero. It happens at the positive value of t such that

$\large -1.8t^2-119.236t+3444.16=0$

solving the quadratic equation we get t = 21.7463 seconds.