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You are an engineer in charge of designing a new generation of elevators for a prospective upgrade to the Empire State Building. Before the state legislature votes on funding for the project, they would like you to prepare a report on the benefits of upgrading the elevators. One of the numbers that they have requested is the time it will take the elevator to go from the ground floor to the 102nd floor observatory. They are unlikely to approve the project unless the new elevators make the trip much faster than the old elevators. If state law mandates that elevators cannot accelerate more than 2.10 m/s2 or travel faster than 10.3 m/s , what is the minimum time in which an elevator can travel the 373 m from the ground floor to the observatory floor

User Dureuill
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1 Answer

3 votes

Answer:

The value is
t_t =41.12 \ s

Step-by-step explanation:

From the question we are told that

The maximum acceleration is
a_m = 2.10 \ m/s^2

The maximum velocity is
v_m = 10.3 \ m/s

The height of the observatory floor is
H = 373 \ m

Generally at the initial state of the motion the elevator accelerates from rest with a maximum acceleration
a_m = 2.10 \ m/s^2 to attain a maximum velocity
a_m = 2.10 \ m/s^2 at a time period of
t_a

Then it continues with the maximum velocity for a time period of
t_b

and finally decelerates with a maximum deceleration
a_m = -2.10 \ m/s^2 to come to rest at a time period of
t_c

Generally the velocity of the elevator for the first part of the motion is


v = u + at_a

given that it started from rest u = 0 m/s

=>
10.3 = 2.10t_a

=>
t_a = 4.90 \ s

Generally from kinematic equations the distance covered during the first part of the motion is


S_1 = ut_a + (1)/(2) at_a^2

=>
S_1 = 0 + (1)/(2) (2.10 ) (4.90)^2

=>
S_1 = 25.21 \ m

Generally the distance covered during the second part of the elevators motion is mathematically represented as


S_2 = v_m * t_2


S_2 = 10.3 t_b

Generally given that the first part of the elevator motion is similar to the third part of the elevator motion it implies that the distance covered is similar

i.e
S_1 = S_3 = 25.21 \ m

The total distance covered is mathematically repented as


H = S_1 + S_2 + S_3

=>
373 = 25.21 + 10.3 t_b + 25.21

=>
t_2 = 31.32 \ s

Generally the minimum total time taken is


t_t = t_a + t_b + t_c

Generally
t_a = t_c = 4.90 \ s

So


t_t  =4.90+ 31.32 +4.90

=>
t_t =41.12 \ s

User Jakub Bujny
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