Hi. So I attached a file of the graph related to this question, on it, the maximum apparent weight of a guy in an elevator is 800N, and it starts on the 50th floor at t=0.00s. I need to know how far the elevator has travelled after 36s (which is the total time on the graph). I'm having a lot of trouble figuring this one out, any ideas?

im having trouble with question too... i tried to break the graph into the 3 sections and figure out the accelerations to then find the distance, but that doesn't work either

somebody help us, plzz!!! thxx

I came up with an answer. Let me know what you get for yours so I can check mine.

What numbers were you getting?

my numbers are different n im pretty sure i did it wrong... if this is a capa question, have u tried the answer?

Originally Posted by Curlyfries

my numbers are different n im pretty sure i did it wrong... if this is a capa question, have u tried the answer?

No, I'm not doing CAPA. I'm past that.

I'm getting 588.6 meters for the question above.
More than half a kilometer does not seem right though.

yah, ur right, the number doesn't seem right... n im totally lost again! i think i might go to the basement of thode tmrw and try to get some help... but if i figure it out before then, i'll post a reply and let you know!

You have to break up the graph into 3 parts according to the ticks on the axes. So the first flat line is 0-6, the large middle section is 6-30s, and the right flat line is 6s.

Also know what the values on the y axis are for each time duration. Use the axis ticks to help you.

1) Find the acceleration for the first 6 seconds by using F=ma, with F being the y-axis value for the middle line minus the y-axis value for the first horizontal line. Basically it's actual weight minus apparent weight. Mass is the mass of the guy

Use d = vt+(1/2)at^2 to find the distance travelled.

Find the final speed at the end of this 6 second duration by using vf^2 = vi^2 + 2ad.
Note that v1 is 0m/s .

2) For the time duration from 6-30s, acceleration is constant bc the guy is moving at a constant velocity. Use d = vt+(1/2)at^2 again to find the distance travelled. Note that the initial speed you use in the equation here is the final speed of the first time duration.

3) Use d = vt+(1/2)at^2 again for the time duration from 30-36s. The speed is the same as the one you used for the middle horizontal line, because the speed didn't change. Acceleration is the negative value of the acceleration you found for the time duration from 0-6s.

Add up all 3 distances. That's the answer.