Question

tion 13 of 19 > 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 3.30 m/s or travel faster than 12.8 m/s, what is the minimum time in which an elevator can travel the 373 m from the ground floor to the observatory floor? minimum travel time: about us careers privacy policy terms of use contact us help 1003 PM 1/28/201 ))

please help

Answer 1

distance to be travelled is 373 m , acceleration _max =3.3 , V _max=12.8

so we when we start at the ground floor we have to start with maximum acceleration and keep accelerating till we reach the maximum allowed speed

so with the given max acceleration and velocity time taken to achieve maximum velocity = v/a=12.8/3.3 = 3.88 sec and the distance traversed in this 3.88 sec is $\frac{1}{2} a t^{2}$ = 0.5*3.3*3.88² = 24.839 m ,then the remaining distance is 373-24.839 = 348.1602 m which is to be travelled with max velocity then the time taken for this segment is s/v = 348.1602/12.8 = 27.2 sec

then the total time will be 3.88+27.2 =31.08 sec

NOTE: If the elevator is to be stopped at the height of the second floor ie,373 m the elevator has to be decelerated to stop at the 102nd floor that means the last 24.839 m will be spent decelerating for 3.88 sec and the distance of 348.1602-24.839=323.3212 m is travelled with maximum velocity then the time taken for the total trip is 3.88(acceleration)+323.3212/12.8(constant max velocity)+3.88(decceleration) =33.0194 sec