If the height of an elevator above the ground is given as h(t)=4t^3-21t^2+18t+30 (h in meters, t>0 sec.) when is acceleration zero?

- Martingale
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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find h''(t) and set it equal to zero and solve for tIf the height of an elevator above the ground is given as h(t)=4t^3-21t^2+18t+30 (h in meters, t>0 sec.) when is acceleration zero?

- stapel_eliz
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
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To expand upon the earlier reply (which is completely correct, by the way), you need to keep in mind that "velocity" is the derivative of "position", and "acceleration" is the derivative of "velocity". So "acceleration" is theIf the height of an elevator above the ground is given as h(t)=4t^3-21t^2+18t+30 (h in meters, t>0 sec.) when is acceleration zero?

That's why you need to find the second derivative, and then set it equal to zero and solve.

i did 12t^2-42t+18= 24t-42=0 t=7/4.is that right?

- Martingale
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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since when does 12t^2-42t+18= 24t-42i did 12t^2-42t+18= 24t-42=0 t=7/4.is that right?

yes...

and

and when

should i use 42/24 insted of 7/4?

- Martingale
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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should i use 42/24 insted of 7/4?

does it matter...they are the same number