Question

Consider a particle moving along the x-axis where
x(t)
is the position of the particle at time t,
x' (t)
is its velocity, and
x'' (t)
is its acceleration.
x(t) = t3 − 12t2 + 21t − 7, 0 ≤ t ≤ 10
(a) Find the velocity and acceleration of the particle.
x' (t) =

x'' (t) =

Answer 1

Find the open t-intervals on which the particle is moving to the right. (Enter your answer using interval notation.)

Answer 2

moving to the right means x is increasing, so x' > 0
x = t^3 − 12t^2 + 21t − 7
x' = 3t^2 - 24t + 21 = 3(t^2-8t+7) = 3(t-1)(t-7)
x" = 6t - 24 = 6(t-4)
so, check where x' > 0 (hint: not between the roots, since it is a parabola which opens up)
Remember the domain.

Answer 3

(0,1)?

Answer 4

no, duh. it is negative between the roots, which are NOT at 0 and 1!