Question

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 m/s21.2 \mathrm{~m} / \mathrm{s}^2, and (2) the typical maximum speed attainable is about 9.0 m/s9.0 \mathrm{~m} / \mathrm{s}. You board an elevator on a skyscraper's ground floor and are transported 180 m180 \mathrm{~m} above the ground level in three steps: acceleration of magnitude 1.2 m/s21.2 \mathrm{~m} / \mathrm{s}^2 from rest to 9.0 m/s9.0 \mathrm{~m} / \mathrm{s}, followed by constant upward velocity of 9.0 m/s9.0 \mathrm{~m} / \mathrm{s}, then deceleration of magnitude 1.2 m/s21.2 \mathrm{~m} / \mathrm{s}^2 from 9.0 m/s9.0 \mathrm{~m} / \mathrm{s} to rest. Find the elapsed time for each of these 33 stages.

Solution

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Answered 1 year ago
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Data known from the problem:

a=1.2ms2a=1.2\hspace{1mm}\frac{\text{m}}{\text{s}^2}

acceleration deceleration is:

ad=1.2ms2a_d=-1.2\hspace{1mm}\frac{\text{m}}{\text{s}^2}

Speed that elevator can reach:

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

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

Elevator first accelerates until it reaches speed vv, 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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