Answer:
A. 2 < t < 4
Explanation:
The position of the particle is given to be:
y = t³ − 6t² + 9t
The velocity of the particle is:
dy/dt = 3t² − 12t + 9
The acceleration of the particle is:
d²y/dt² = 6t − 12
Find when the acceleration is 0:
0 = 6t − 12
t = 2
In the interval 0 < t < 2, the acceleration is negative (the particle's velocity is decreasing).
In the interval 2 < t < 4, the acceleration is positive (the particle's velocity is increasing).