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
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stellar wrote: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?

find h''(t) and set it equal to zero and solve for t

- stapel_eliz
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stellar wrote: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?

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 the

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

stellar wrote: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?

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

- Martingale
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stellar wrote:

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

since when does 12t^2-42t+18= 24t-42

yes...

and

and when

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

- Martingale
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stellar wrote:should i use 42/24 insted of 7/4?

does it matter...they are the same number