Integration of Acceleration to get the height of a building

In summary: The variables here are s (distance), u (initial velocity), t (time), and a (acceleration). In summary, the conversation is about a University Physics lab where the goal is to find the height of a building using weight and time measurements in an elevator. The equation a/g=(W-W_{o})/W_{o} is mentioned and it is known that it needs to be integrated twice to find the height. There is confusion about how to integrate with numerical variables and the user is seeking help with the problem.
  • #1
kli53
1
0

Homework Statement



Thank you first off for your help. Currently, I am working on a lab for my University Physics class and I am lost on how to do this integration. We weight (from a scale) and time measurements going up and down in an elevator. Using that data we are supposed to get the height of the building (in our case floors 1-4).

1st Run
1st floor (acceleration up) 2.4 second 280 pounds 16.8 seconds
4th Floor (deceleration up) 3.32 seconds 255 pounds

2nd Run
4th floor (acceleration down) 2.16 seconds 280 pounds 23 seconds
1st floor (deceleration down) 3.13 seconds 255 pounds

3rd Run
1st floor (acceleration up) 2.56 seconds 260 pounds 14.4 seconds
4th Floor (deceleration up) 2.06 seconds 255 pounds

4th Run
4th floor (acceleration down) 4.3 seconds 255 pounds 22.86 seconds
1st floor (deceleration down) 3.26 seconds 260 pounds

Homework Equations



a/g=(W-W[itex]_{o}[/itex])/W[itex]_{o}[/itex]

The Attempt at a Solution



None yet, though I am assuming we are integrating to time. But I am confused, since all of my variables are numbers. So if you have a=dv/dt, how would that work with this equation?I know this equation needs to be integrated twice to get the height, however, I am just not sure where to start with that. I have taken Calc 3, but this for some reason just confuses me.

Thank you again for your time!
 
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  • #2
kli53 said:
1st floor (acceleration up) 2.4 second 280 pounds 16.8 seconds
4th Floor (deceleration up) 3.32 seconds 255 pounds
How to interpret this?
A body in the lift weighed 290 lbs for 2.4 secs, then X lbs for 16.8 secs, then 255 lbs for 3.32 secs? Is that correct? Do you know what X is?

A equation relating distance to acceleration is s = ut + ½at2
 

1. How does integrating acceleration help determine the height of a building?

Integrating acceleration is a mathematical process that allows us to find the distance traveled over a certain period of time. By integrating the acceleration of an object falling due to gravity, we can determine the height of a building it was dropped from.

2. What are the steps involved in integrating acceleration to find the height of a building?

The first step is to measure the initial velocity and acceleration of the falling object. Then, using the equation d = (1/2)at^2 + vt, where d is the distance traveled, a is acceleration, t is time, and v is initial velocity, we can calculate the height of the building.

3. Is integrating acceleration a reliable method for determining the height of a building?

Integrating acceleration can provide a fairly accurate estimate of the height of a building, but it is not always 100% reliable. Factors such as air resistance and wind can affect the accuracy of the calculation.

4. Can integrating acceleration be used for any type of building?

Integrating acceleration can be used for any type of building, as long as the building has a known initial velocity and acceleration. However, it may not be as accurate for buildings with unusual shapes or structures.

5. Are there any other methods for determining the height of a building besides integrating acceleration?

Yes, there are other methods such as using a laser rangefinder or measuring the shadow of the building at a specific time of day. However, integrating acceleration is a common and efficient method for determining height in many scientific and engineering applications.

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