A car is to be hoisted by elevator to the fourth floor of a parking garage, which is at a heigh 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.

Classical Dynamics of Particles and Systems
5th Edition
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Stephen T. Thornton, Jerry B. Marion
Chapter2: Newtonian Mechanics-single Particle
Section: Chapter Questions
Problem 2.16P: A particle is projected with an initial velocity 0 up a slope that makes an angle with the...
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Q18

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 az, and reach a max imum
speed v, determine the shortest time to make the lift, starting from rest and ending at rest.
m
h = 16 m
aj = 0.2
az = 0.1
2
v = 2.7
Given:
S
Solution:
Assume that the elevator never reaches its maximum speed.
m
tj = 1 s
12 = 2 s
hị = 1 m
Guesses
Vmax = 1
Given
Vmax = ajti
1
2
0 = vmax - at2 - t)
h = hj + Vualt2 - 1) - ;az(t2 - 4)²
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 az, and reach a max imum speed v, determine the shortest time to make the lift, starting from rest and ending at rest. m h = 16 m aj = 0.2 az = 0.1 2 v = 2.7 Given: S Solution: Assume that the elevator never reaches its maximum speed. m tj = 1 s 12 = 2 s hị = 1 m Guesses Vmax = 1 Given Vmax = ajti 1 2 0 = vmax - at2 - t) h = hj + Vualt2 - 1) - ;az(t2 - 4)²
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