Given the position function, s of t equals t cubed divided by 3 minus 12 times t squared divided by 2 plus 36 times t, between t = 0 and t = 15, 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.

Answer 1

s(t) = t^3/3 - 12t^2/2 + 36t
Strange, saying 12t^2/2 rather than 6t^2, but hey ...

Assuming "moving to the right" means s(t) is increasing, then we just have to find when ds/dt > 0. That is,

t^2 - 12t + 36 > 0

Hmmm. Something tells me you had a typo, and it was really

s(t) = t^3/3 - 13t^2/2 + 36t
Then
ds/dt = t^2 - 13t + 36 > 0
(t-9)(t-4) > 0

Now I suspect another typo, because ds/dt > 0 when t<4 or t>9.

Fix your mistakes and follow the logic to a correct solution.

Answer 2

Steve, you idiot. There were no typos you completely ignorant buffoon. This person copied and pasted the problem so don't suspect they made a typo and don't assume they made a mistake you cocky peon.