Position, Acceleration, and Velocity of an Elevator

In summary, the problem involves a hotel elevator that ascends 200 m with a maximum speed of 4.8 m/s, and its acceleration and deceleration both have a magnitude of 1.0 m/s2. The first part of the problem requires finding the distance the elevator moves while accelerating to full speed from rest, which can be solved using the formula x = 1/2 a*t2. The second part involves determining the time it takes to make the complete trip from bottom to top, which can be found by dividing the maximum speed by the acceleration.
  • #1
Phoenixtears
83
0

Homework Statement


A hotel elevator ascends 200 m with a maximum speed of 4.8 m/s. Its acceleration and deceleration both have a magnitude of 1.0 m/s2.

(a) How far does the elevator move while accelerating to full speed from rest? (Answer in Meters)

(b) How long does it take to make the complete trip from bottom to top? (Answer in seconds)

Homework Equations


Velocity= delta-x/ delta-t

Acceleration= Delta-v/ delta-t


The Attempt at a Solution



For some reason I can't reason this one out. I have attempted to graph the velocity vs. time graph (which I believe will hold the solutions) and my graph turned out to be a triangle with a little below the x-axis. Yet this graph doesn't look quite right to me, and I can't figure out how to derive the correct velocity vs. time graph from the given information. I have also tried to draw the acceleration graph, slowly speeding up the elevator (1 m/s^2), but my graph has almost leveled out, proving something must be entirely wrong. Pure math isn't doing it for me. Any ideas?
 
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  • #2
Phoenixtears said:

Homework Statement


A hotel elevator ascends 200 m with a maximum speed of 4.8 m/s. Its acceleration and deceleration both have a magnitude of 1.0 m/s2.

(a) How far does the elevator move while accelerating to full speed from rest? (Answer in Meters)

(b) How long does it take to make the complete trip from bottom to top? (Answer in seconds)

Homework Equations


Velocity= delta-x/ delta-t
Acceleration= Delta-v/ delta-t

The Attempt at a Solution



For some reason I can't reason this one out. I have attempted to graph the velocity vs. time graph (which I believe will hold the solutions) and my graph turned out to be a triangle with a little below the x-axis. Yet this graph doesn't look quite right to me, and I can't figure out how to derive the correct velocity vs. time graph from the given information. I have also tried to draw the acceleration graph, slowly speeding up the elevator (1 m/s^2), but my graph has almost leveled out, proving something must be entirely wrong. Pure math isn't doing it for me. Any ideas?

Maybe break it up into several problems? How long does it take to accelerate to the Vmax. That's almost the first part of the problem you have to find anyway. Happily it takes the same time to slow down at the same rate.

Once you found the time to speed up you can find the distance to speed up. That's the answer to part A.The distance to slow down is again happily the same. Then you know the distance that you will travel at Vmax and hence the time.

The formulas here may also help you:
https://www.physicsforums.com/showpost.php?p=905663&postcount=2
 
  • #3
LowlyPion said:
Maybe break it up into several problems? How long does it take to accelerate to the Vmax. That's almost the first part of the problem you have to find anyway. Happily it takes the same time to slow down at the same rate.

That really seemed like a perfect way to attack the problem after I read your post, but I can't figure out how long it takes to accelerate. I mean, I think the wording is confusing me. Do they mean the is takes just 1 m/s^2 to accelerate to 4.8 m/s? And then, how can I translate m/s^2 into velocity? Is that even possible with just the number 1? I'm completely flabbergasted on this one. Any other suggestions?
 
  • #4
Phoenixtears said:
That really seemed like a perfect way to attack the problem after I read your post, but I can't figure out how long it takes to accelerate. I mean, I think the wording is confusing me. Do they mean the is takes just 1 m/s^2 to accelerate to 4.8 m/s? And then, how can I translate m/s^2 into velocity? Is that even possible with just the number 1? I'm completely flabbergasted on this one. Any other suggestions?

The statement of the problem tells you the acceleration. Velocity from 0 is time * acceleration. Hence time is Velocity / acceleration. Time to max speed then it 4.8 / 1.

Distance is related by the equation x = 1/2 a*t2

These equations are at the link I provided.
 

1. What is the difference between position, acceleration, and velocity of an elevator?

Position refers to the location of the elevator in relation to a fixed point. Acceleration is the rate of change of velocity over time, and velocity is the speed and direction of the elevator's movement.

2. How are position, acceleration, and velocity related to each other?

Position is the integral of velocity, and velocity is the integral of acceleration. This means that the change in acceleration affects the velocity, and the change in velocity affects the position.

3. How are the position, acceleration, and velocity of an elevator measured?

The position of an elevator can be measured using a sensor or by tracking the movement of the elevator over time. Acceleration and velocity can be measured using accelerometers or by calculating the change in position over time.

4. How does the position, acceleration, and velocity of an elevator change during its journey?

At the start of its journey, the elevator's position is at a fixed point and its acceleration and velocity are both zero. As the elevator starts moving, its acceleration increases and velocity also starts to increase. Once the elevator reaches its desired speed, the acceleration becomes constant and the velocity remains constant until the elevator reaches its destination. At this point, the acceleration decreases to zero and the elevator's velocity decreases to zero as it comes to a stop at the final position.

5. How do factors such as weight and friction affect the position, acceleration, and velocity of an elevator?

The weight of the elevator and its passengers affects the force needed to accelerate and decelerate the elevator, which in turn affects its acceleration and velocity. Friction can also impact the movement of the elevator, causing it to require more force to move and slowing down its acceleration. Additionally, the weight and friction can also affect the maximum velocity that the elevator can reach.

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