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

[mossfan563]

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?

 

[Doc Al]

What have you done so far? How quickly does it reach maximum speed? What distance does that require?

 

[mossfan563]

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.

 

[Doc Al]

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.

 

[mossfan563]

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.

 

[LowlyPion]

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.

 

[mossfan563]

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.

 

[LowlyPion]

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

 

[mossfan563]

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?

 

[LowlyPion]

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.