## Related questions with answers

High-speed elevators function under two limitations: (1) the maximum magnitude of vertical acceleration that a typical human body can experience without discomfort is about $1.2 \mathrm{~m} / \mathrm{s}^2$, and (2) the typical maximum speed attainable is about $9.0 \mathrm{~m} / \mathrm{s}$. You board an elevator on a skyscraper's ground floor and are transported $180 \mathrm{~m}$ above the ground level in three steps: acceleration of magnitude $1.2 \mathrm{~m} / \mathrm{s}^2$ from rest to $9.0 \mathrm{~m} / \mathrm{s}$, followed by constant upward velocity of $9.0 \mathrm{~m} / \mathrm{s}$, then deceleration of magnitude $1.2 \mathrm{~m} / \mathrm{s}^2$ from $9.0 \mathrm{~m} / \mathrm{s}$ to rest. Find the elapsed time for each of these $3$ stages.

Solution

VerifiedData known from the problem:

$a=1.2\hspace{1mm}\frac{\text{m}}{\text{s}^2}$

acceleration deceleration is:

$a_d=-1.2\hspace{1mm}\frac{\text{m}}{\text{s}^2}$

Speed that elevator can reach:

$v=9\hspace{1mm}\frac{\text{m}}{\text{s}}$

$y=180\hspace{1mm}\text{m}$

Elevator first accelerates until it reaches speed $v$, than it is moving with constant speed, and than decelerates until it stops. We need to find time needed for each interval.

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