Mark M. answered 12/14/23
Retired math prof. Very extensive Precalculus tutoring experience.
v(t) = ∫a(t)dt = ∫(-6)dt = -6t + C
Since v(2) = -27, we have -12 + C = -27. So, C = -15
v(t) = -6t - 15
s(t) = ∫v(t)dt = ∫(-6t - 15)dt = -3t2 - 15t + C1
Since s(1) = 8, -18 + C1 = 8. So, C1 = 26.
s(t) = -3t2 - 15t + 26