Physics like antiderivative problem...

[des4ij]

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 -18 ft/s in 5 s. The rocket then "floats" to the ground at that rate.

At what time does the rocket reach its maximum height?
What is that height?

At what time does the rocket land?

thx for any help... I think it may be done in like a lolng way taking each part and integrating each part but i am still confused how to set the problem up...

 

[skeeter]

you'll have to do this problem in stages ...

for 0 < t < 3 ...

a(t) = 60t
v(t) = 30t² + C
assuming v(0) = 0 ... C = 0
v(t) = 30t²
h(t) = 10t³ + C
once again, assuming h(0) = 0 ... C = 0
h(t) = 10t³

at t = 3 s, h(3) = 270 ft, v(3) = 270 ft/s


for 3 < t < 17 ...

a(t) = -32
v(t) = -32t + C
since v(3) = 270 ... C = 366
v(t) = -32t + 366
h(t) = -16t² + 366t + C
since h(3) = 270 ... C = -684
h(t) = -16t² + 366t - 684

at t = 17 s, v(17) = -178 ft/s, h(17) = 914 ft

for 17 < t < 22 ...

the speed decreases from -178 ft/s to -18 ft/s ... a = +32 ft/s²

a(t) = 32
v(t) = 32t + C
since v(17) = -178 ... C = -722
v(t) = 32t - 722
h(t) = 16t² - 722t + C
since h(17) = 914 ... C = 8564
h(t) = 16t² - 722t + 8564

at t = 22 s, v(22) = -18 ft/s, h(t) = 424 ft

for t > 22 until it hits the ground, v(t) = -18 ft/s, a constant.

you should be able to take over and answer all necessary questions.

hope I didn't mess up somewhere along the way, so check my work

... neat problem!