Question

How do you do this question?

Answer 1

Answer:

A. 2 < t < 4

Step-by-step 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).