Question

A car is to be hoisted by elevator to the fourth floor of a parking garage, which is 48 ft above the ground. If the elevator can accelerate at 2 0.6 / ft s ,decelerate at 2 0.3 / ft s , and reach a maximum speed of 8 / ft s, determine the shortest time to make the lift, starting from rest and ending at rest. t s  21.9

Answer 1

Answer: 21.91 s

Explanation:

Given that,

Maximum height of the car, h = 48 ft

Acceleration of the elevator, a = 0.6 ft/s²

Deceleration of the elevator, -a = 0.3 ft/s²

Maximum speed of the elevator, v = 8 ft/s

Initial speed of the elevator, u = 0

If when the elevator accelerate from 0 to maximum velocity, v.

Let s be the vertical distance traveled during acceleration.

v² = u² - 2as

s = (v² - u²) / 2a

s = (8² - 0) / 2*0.6

s = 64 / 1.2

s = 53.33 ft

If when the elevator decelerates from maximum velocity, v to zero.

Let S be the vertical distance traveled during deceleration

u² = v² + 2aS

S = (u² - v²) / 2a

S = (0 - 8²) / 2 * 0.3

S = -64 / 0.6

S = 106.67 ft

Since he sum of s and S (i.e s + S) is greater than 48 ft, then the elevator will switch from acceleration to deceleration

without reaching the maximum velocity. Below, the switching point is labeled y.

v² = u² + 2ay

y = v²/2a

Inserting this into the earlier deceleration equation, we have

-v²/2 = d * [48 - (v²/2a)], where

d = deceleration

a = acceleration

Therefore, v = [4.√6. a √-(a.b/a)] / b

Where b = acceleration - deceleration

v = 4.382 ft/s

Using this newly found v, we proceed to find our s

s = (u² + v²)/2a

s = 19.2 / 1.2

s = 16 ft

The transport times for each segment are found from

v = u + a*t, thus upward t1

4.382 = 0 + 0.6 * t

t = 4.382/0.6

t = 7.303 s

Also,

4.382 = 0 + 0.3 * T

T = 4.382/0.3

T = 14.607 s

The total travel time is then t + T =

7.303 + 14.607

Total time of travel is 21.91 s