anti-derivative help PLEASE!!!?

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. 14 seconds later, the rocket's parachute opens, and the (downward) velocity slows linearly to -18ft/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"). 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?

Anonymous · Accepted Answer

I can't sketch the problem for you, but I can work it out! [For you to understand!]

(A) Velocity function = V(T)= Integral of 60t= 60t^2/2= 30(t^2)
  Position = S(T)= Integral of 30(t^2) =30 [T^3/3] =10 (t^3)

(B)  <using a sign graph> The rocket will reach its maximum height when V(T) changes from Positive to Negative, or at endpoints of the function (in this case, 3) . 

V(T) =0 ;;;;; 30(t^2)=0 and T=0; when T=0, position=0. We can invalidate T=0 as an absolute max. 

Test S(3). S(T)= 10 (t^3); S(3)= 10 (3^3)= 10 (27) = 270 feet.
At T=3 seconds, the rocket reaches 270 feet, which is its maximum height.

(C) I don't understand this, but I would use the graph of V(T). You would have to make two integrals, one from 0 to 3, and then from that point (where it changes direction), do an integral from when it goes down all the way to when it ends.

*edit: i searched this problem on yahoo and someone asked the same thing 5 months ago. http://answers.yahoo.com/question/index?qid=20071028100422AAWdqXl 
Although i think the explanation in that question was a bit complex, i'm not sure whether he was right or not

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anti-derivative help PLEASE!!!?

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