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und. If the elevator can accelerate at a,, decelerate at a, and reach a ma mine the shortest time to make the lift, starting from rest and ending at 1 h = 16 m m a, = 0.2 m az = 0.1 = v = 2.7

und. If the elevator can accelerate at a,, decelerate at a, and reach a ma mine the shortest time to make the lift, starting from rest and ending at 1 h = 16 m m a, = 0.2 m az = 0.1 = v = 2.7

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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A car is to be hoisted by elevator to the fourth floor of a parking garage, which is at a height h above the ground. If the elevator can accelerate at a,, decelerate at a, and reach a max imum speed v, determine the shortest time to make the lift, starting from rest and ending at rest. m m h = 16 m aj = 0.2 az = 0.1 v = 2.7 Given: S Solution: Assume that the elevator never reaches its maximum specd. h = 1 m m 11 = 1 s 12 = 2 s Guesses Vmax Given Vmax = ajti h = 0 = vmax – a{(12 - t) h = h1 + Vua(42 – 1) - a2lt2 - 4)²
Transcribed Image Text:A car is to be hoisted by elevator to the fourth floor of a parking garage, which is at a height h above the ground. If the elevator can accelerate at a,, decelerate at a, and reach a max imum speed v, determine the shortest time to make the lift, starting from rest and ending at rest. m m h = 16 m aj = 0.2 az = 0.1 v = 2.7 Given: S Solution: Assume that the elevator never reaches its maximum specd. h = 1 m m 11 = 1 s 12 = 2 s Guesses Vmax Given Vmax = ajti h = 0 = vmax – a{(12 - t) h = h1 + Vua(42 – 1) - a2lt2 - 4)²
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