x(t)=3t^3-40.5t^2+162t find where object is speeding up or slowing down

Answer 1

x ' (t) = 9t^2 - 81t + 162
so objects are "speeding up" when x ' (t) > 0
and slowing down when x ' (t) < 0
Consider the zeros of the expression
9t^2 - 81t + 162 = 0
t^2 - 9t + 18 = 0
(t-3)(t-6) = 0
t = 3 or t = 6
so the t of the vertex would be half-way between those values or t = 4.5

Using the properties of the quadratic and knowing that x ' (t) is a parabola opening upwards
the object is speeding up for t > 4.5 seconds
and is slowing down for t < 4.5 seconds

Here is a graph of the velocity
http://www.wolframalpha.com/input/?i=plot+y+%3D+9t%5E2+-+81t+%2B+162