# calculus

Find the position function s(t) given acceleration a(t)=3t if v(2)=0 and s(2)=1.

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1. a(t) = 3t
v(t) = (3/2)t^2 + c
v(2) = 0 ----> 0 = (3/2)(4) + c
0 = 6 + c
c = -6
v(t) = (3/2)t^2 - 6
s(t) = (1/2)t^3 - 6t + k
s(2) = 1 -----> 1 = 4 - 12 + k
k = 9

s(t) = (1/2)t^3 - 6t + 9

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posted by Reiny

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