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