per the usual rules,
(a) v(t) = 6t^2-6t-30 = 6(t^2-t-5)
(b) a(t) = 6(2t-1)
(c) solve for v=0 and then evaluate s there
(d) is v(6) positive or negative?
The position of a point on a line is given by the equation s(t)= 2t^3-3t^2-30t+36 where s is measured in metres and t in seconds.
a) what is the velocity of the point after 2 seconds?
b) what is its acceleration after 3 seconds?
c) where is it when it first stops moving?
d) After 6 seconds, is it moving toward or moving away from the origin?
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