# Integration of Acceleration to get the height of a building

1. Feb 5, 2012

### kli53

1. The problem statement, all variables and given/known data

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

2. Relevant equations

a/g=(W-W$_{o}$)/W$_{o}$

3. 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!

Last edited: Feb 5, 2012
2. Feb 5, 2012

### Staff: Mentor

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