## height h(t)=4t^3-21t^2+18t+30: when is acceleration zero?

Limits, differentiation, related rates, integration, trig integrals, etc.
stellar
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### height h(t)=4t^3-21t^2+18t+30: when is acceleration zero?

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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### Re: height h(t)=4t^3-21t^2+18t+30: when is acceleration zero?

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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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 second derivative of "position".

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

stellar
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### Re: height h(t)=4t^3-21t^2+18t+30: when is acceleration zero?

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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### Re: height h(t)=4t^3-21t^2+18t+30: when is acceleration zero?

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

yes...$v(t)=h'(t)=12t^2-42t+18$

and $a(t)=v'(t)=h''(t)=24t-42$

and $a(t)=0$ when $t=\frac{42}{24}$

stellar
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### Re: height h(t)=4t^3-21t^2+18t+30: when is acceleration zero?

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

Martingale
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### Re: height h(t)=4t^3-21t^2+18t+30: when is acceleration zero?

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

does it matter...they are the same number