A particle moves along a vertical line. Its position in m at time t in s is given by s(t) = t3 – 9t2 + 24.
a) Determine the distance the particle travels between t = 0 & t = 6
b) When is the particle speeding up?
(a) The particle is moving up when s'(t) > 0
s'(t) = 3t^2 - 18t = 3t(t-6)
Note that s'(t) < 0 on [0,6]
That means the particle starts at +24m and finishes at -84m
It moved down 100m
(b) speeding up when velocity increases
s"(t) = 6t-18 = 6(t-3)
s"(t) > 0 on [3,6]
That is when it has positive acceleration -- it is speeding up.