How Long Does It Take an Elevator to Travel 199 Meters from Start to Stop?

In summary: Going 174 m at 5 m/sec should be accomplished considerably quicker if you managed to slow up and down in only 9 seconds over 25 feet.
  • #1
mossfan563
54
0
A certain elevator cab has a total run of 199 m and a maximum speed is 313 m/min, and it accelerates from rest and then back to rest at 1.10 m/s2. (b) How long does it take to make the nonstop 199 m run, starting and ending at rest?

How do I approach part B?
 
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  • #2
What have you done so far? How quickly does it reach maximum speed? What distance does that require?
 
  • #3
Doc Al said:
What have you done so far? How quickly does it reach maximum speed? What distance does that require?

Well part A was how far does the cab move while accelerating to full speed from rest?
And the answer was 12.3698 m.
That's what I've done so far.
 
  • #4
mossfan563 said:
Well part A was how far does the cab move while accelerating to full speed from rest?
And the answer was 12.3698 m.
That's what I've done so far.
Good.

Treat the motion as having three stages: Accelerating to max speed; constant speed; decelerating to zero speed.

Figure out the distance and time associated with each stage of the motion.
 
  • #5
Doc Al said:
Good.

Treat the motion as having three stages: Accelerating to max speed; constant speed; decelerating to zero speed.

Figure out the distance and time associated with each stage of the motion.

Well I technically have the distance and time for accelerating to max speed. I can't figure out the other two though. I tried the quadratic equation and it doesn't work.
 
  • #6
mossfan563 said:
Well I technically have the distance and time for accelerating to max speed. I can't figure out the other two though. I tried the quadratic equation and it doesn't work.

What quadratic equation? You have 3 regions. There are the two - one at each end - getting to max speed and going back to 0, and you know the distances for both because they are symmetrically the same.

Since you know the distance for both, then you know what remains of the trip that you will account for with max speed. Knowing the distance and knowing the max speed then you can figure the time.

Time for the trip? Total time = Time to speed + time at max + time to slow to 0.
 
  • #7
LowlyPion said:
What quadratic equation? You have 3 regions. There are the two - one at each end - getting to max speed and going back to 0, and you know the distances for both because they are symmetrically the same.

Since you know the distance for both, then you know what remains of the trip that you will account for with max speed. Knowing the distance and knowing the max speed then you can figure the time.

Time for the trip? Total time = Time to speed + time at max + time to slow to 0.

Ok I understand that the time to speed and time to slow to 0 are pretty much the same. The time at max still is still confusing me. The remaining distance would be around 174.2.
I do know the max speed. I just don't know what formula to use.
 
  • #8
mossfan563 said:
Ok I understand that the time to speed and time to slow to 0 are pretty much the same. The time at max still is still confusing me. The remaining distance would be around 174.2.
I do know the max speed. I just don't know what formula to use.

V= x/t
 
  • #9
Ok I used this formula.
Here are the numbers I used.
Total distance = 199 m
Remaining distance = 174.2604 m
Time of speed up and slow down = 4.742... s
Max speed = 5.21666666... m/s

Using all these numbers I got an answer of 919.49 s.
Is this off? What am I doing wrong? Are my numbers wrong?
 
  • #10
mossfan563 said:
Ok I used this formula.
Here are the numbers I used.
Total distance = 199 m
Remaining distance = 174.2604 m
Time of speed up and slow down = 4.742... s
Max speed = 5.21666666... m/s

Using all these numbers I got an answer of 919.49 s.
Is this off? What am I doing wrong? Are my numbers wrong?

Going 174 m at 5 m/sec should be accomplished considerably quicker if you managed to slow up and down in only 9 seconds over 25 feet.
 

Related to How Long Does It Take an Elevator to Travel 199 Meters from Start to Stop?

1. What is the basic principle behind elevator operation?

The basic principle behind elevator operation is the concept of counterweights and pulleys. The elevator car is attached to a system of cables and pulleys, which are connected to a counterweight. The weight of the counterweight is equal to the weight of the elevator car and its maximum capacity. This allows the elevator to move up and down smoothly, as the weight of the car is balanced by the counterweight.

2. How does an elevator know when to stop at a certain floor?

Modern elevators are equipped with sensors and a control system that detects the position of the elevator car in the shaft. When a passenger presses a button for a specific floor, the control system calculates the distance the elevator needs to travel and sends a signal to the motor to stop at the desired floor.

3. What is the difference between an electric and hydraulic elevator?

An electric elevator uses an electric motor and a system of cables and pulleys to move the elevator car. A hydraulic elevator, on the other hand, uses a hydraulic fluid and a piston to move the elevator. Electric elevators are more commonly used in high-rise buildings, while hydraulic elevators are more suitable for low-rise buildings.

4. How do elevators prevent accidents and ensure safety?

Elevators are equipped with several safety features, such as emergency brakes, door sensors, and a system that prevents the elevator from moving unless all doors are securely closed. Additionally, elevators are regularly inspected and maintained by trained professionals to ensure their safety and prevent accidents.

5. Why do elevators sometimes make unexpected stops or movements?

Elevators may make unexpected stops or movements due to various reasons, such as a malfunctioning sensor, a power outage, or a passenger pressing the wrong button. In such cases, the elevator's safety features will activate and bring the elevator to a stop to prevent any accidents.

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