v(t)= t^3 - 3t^2 + 3t
Are there any times after t = 0 where the particle comes to a stop?
Does the particle ever slow down?
At a stop, the velocity is zero, so
t^3 - 3t^2 + 3t = 0
t(t^2 - 3t + 3 ) =0
t = 0 or t^2 -3t + 3 = 0
t^2 - 3t + 3 = 0 has no real solution
to slow down a(t) would have to be negative
a(t) = 3t^2 - 6t + 3
let's see where it is zero
t^2 - 2t + 1 = 0 , after dividing both sides by 3
(t-1)^2 =
t = 1
let's look at the graphs
http://www.wolframalpha.com/input/?i=plot+y+%3D+t%5E3+-+3t%5E2+%2B+3t+%2C+y+%3D+3t%5E2+-+6t+%2B+3
the blue graph is the velocity
the red graph is the accelertion
since the red graph is decreasing from 0 to 1, the acceleration is decreasing between t = 0 and t = 1, so the particle is slowing down for that time interval
notice that the blue graph , the velocity, crosses the t axis only at t = 0, so it never comes to a stop other than the starting time of t = 0