Question

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

Answer 1

Answer:

The value is $t_t =41.12 \ s$

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

$S_1 = ut_a + \frac{1}{2} at_a^2$

=> $S_1 = 0 + \frac{1}{2} (2.10 ) (4.90)^2$

=> $S_1 = 25.21 \ m$

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

     $t_t =4.90+ 31.32 +4.90$

=> $t_t =41.12 \ s$