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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. Given: h = 16 m m m aj = 0.2 az = 0.1 - v = 2.7

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. Given: h = 16 m m m aj = 0.2 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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Question 18

Problem 12–18 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 2 v = 2.7 Given: Solution: Assume that the elevator never reaches its maximum speed. m 1 s t2 = 2 s Vmax = 1 S hị = 1 m Guesses t1 Given Vmax = ajti 1 h1 0 = Vmax h = hị + Ynalt2 - ti) - azlt2 - 1)*
Transcribed Image Text:Problem 12–18 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 2 v = 2.7 Given: Solution: Assume that the elevator never reaches its maximum speed. m 1 s t2 = 2 s Vmax = 1 S hị = 1 m Guesses t1 Given Vmax = ajti 1 h1 0 = Vmax h = hị + Ynalt2 - ti) - azlt2 - 1)*
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