Question

Motion and rate of change derivatives. Attached is the question and where I am stuck, I would really appreciate the help :)

Answer 1

Answer:

A. 3 < t < 9

Step-by-step explanation:

When the particle is moving to the right, the velocity is positive.

To find the velocity, take the derivative of s(t) with respect to t.

s(t) = -t³/3 + 13t²/2 − 30t

s'(t) = -t² + 13t − 30

Find when the velocity is 0.

0 = -t² + 13t − 30

0 = t² − 13t + 30

0 = (t − 3) (t − 10)

t = 3, 10

Check the sign of s'(t) in each interval.

0 < t < 3, s'(t) < 0

3 < t < 9, s'(t) > 0

The particle moves to the right when 3 < t < 9.