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
where C is a constant.
But v(0) = 0 (the rocket is launched from rest), so C = 0 and
S(t) is the anti-derivative of v(t), so
where C is a constant.
But s(0) = 0 (the rocket is on the land), so C = 0 and
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
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
But
Hence
and
for 3≤ t ≤ 17
At second 17 the parachute opens and the rocket gets the acceleration of
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
since
then
and
until landing.
When the rocket lands, s(t) becomes zero. It happens at the positive value of t such that
solving the quadratic equation we get t = 21.7463 seconds.
So
for 17 ≤ t ≤ 21.7463
Summarizing
and
(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)
(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
solving the quadratic equation we get t = 21.7463 seconds.
