A particle moves according to a law of motion s = f(t) = t3 - 12t2 + 21t, t 0, where t is measured in seconds and s in feet.

Question

v(t) = 3t^2 - 24t 21
a(t) = 6t - 24

e) Find the total distance traveled during the first 8 s.

i) When is the particle speeding up? (Enter your answers in ascending order. If you need to use -∞ or ∞, enter -INFINITY or INFINITY.)

( , ) U ( ,infinity)

When is it slowing down?
(0, ) U ( , )

Answer 1

e) Substitute t = 8s into
S = t^3 -12t^2 + 21t

i) Speeding up = acceleration > 0
when is a > 0?
It's when 6t - 24 > 0

Slowing down = acceleration < 0

(just as extra info: constant speed would be when your acceleration = 0)