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An elevator cab in the New York Marriott Marquis (see Fig. 2-39) has a total run of 624 ft. Its maximum speed is 1000 ft/min and its (constant) acceleration is 4.00 ft/s'', (a) How far does it move while accelerating to full speed from rest? (b) How long does it take to make the run, starting and ending at rest?

An elevator cab in the New York Marriott Marquis (see Fig. 2-39) has a total run of 624 ft. Its maximum speed is 1000 ft/min and its (constant) acceleration is 4.00 ft/s'', (a)
How far does it move while accelerating to full speed from rest? (b) How long does it take to make the run, starting and ending at rest?

Grade:upto college level

1 Answers

Navjyot Kalra
askIITians Faculty 654 Points
6 years ago
Given:
Initial speed of the elevator cab, V0 = 0 ft/s.
Final speed of the elevator cab, v = 1000 ft / min.
Acceleration of elevator cab, a = 4.00 ft/s2.
Let us assume that the time taken by the elevator cab to attain its maximum speed from rest is given by t.
First, convert the final speed of the elevator cab from ft / min to ft/s as:
232-2391_1.PNG
Therefore the final speed of the elevator cab is 16.6 ft/s.
From the equation of kinematics, we have
232-137_2.PNG
232-1612_3.PNG
We assume that the distance travelled by the elevator cab when it starts from rest and finish at rest is given by s, such that the length of the elevator total run is equal to s.Given:
Length of elevator’s total run, s = 624 ft.
Initial speed of the elevator, v0 = 0 ft /s.
The distance travelled by the elevator during the total run is given as:
232-33_4.PNG
From the assumption it is clear that s0 = 0 ft.
Therefore substitute the given values and appropriate initial conditions in the equation above to have
232-1297_5.PNG
It is important to note that the negative solution of time is discarded.
Therefore the time taken by the elevator cab to travel its entire run, such that it begins initially from rest and stops at rest is 17.6 s.

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