Asked by Mel
Given the position function, s of t equals negative t cubed divided by 3 plus 13 times t squared divided by 2 minus 30 times t, between t = 0 and t = 9, where s is given in feet and t is measured in seconds, find the interval in seconds where the particle is moving to the right.
3 < t < 9
5 < t < 9
The particle never moves to the right.
The particle always moves to the right. <- This is my answer
3 < t < 9
5 < t < 9
The particle never moves to the right.
The particle always moves to the right. <- This is my answer
Answers
Answered by
Steve
what's with all the words?
s(t) = -t^3/3 + 13t^2/2 - 30t
when the particle is mofing to the right, s(t) is increasing, so ds/dt > 0
-t^2 + 13t - 30 > 0
3 < t < 10
Since our domain is 0<t<9, our interval is thus
3<t<9
s(t) = -t^3/3 + 13t^2/2 - 30t
when the particle is mofing to the right, s(t) is increasing, so ds/dt > 0
-t^2 + 13t - 30 > 0
3 < t < 10
Since our domain is 0<t<9, our interval is thus
3<t<9
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