Question 626996
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Your problem doesn't specify the speed when the elevator is going down, and there is no reason to assume that the speed going up and the speed going down is the same, hence the following function is only valid for modeling the situation where the elevator is going up.


Presuming that a miner could enter or exit the elevator at intermediate points between the bottom of the shaft and the surface, a start depth, *[tex \LARGE h_o] must be specified.


The distance traveled in *[tex \LARGE t] seconds is *[tex \LARGE 20t] feet since the elevator rises at 20 feet per second.


Taking into account that *[tex \LARGE h_o] is a distance below ground and is therefore a negative number, we write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ 20t\ +\ h_o]


where


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -12720\ \leq\ h_o\ \leq\ 0]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \text{dom}(h)\ =\ \left\{t\ \in\ \mathbb{R}\ |\ 0\ \leq\ t\ \leq\ \frac{h_o}{20}\right\}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \text{ran}(h)\ =\ \left\{h(t)\ \in\ \mathbb{R}\ |\ h_o\ \leq\ h(t)\ \leq\ 0\right\}]





John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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