Recent content by warsno

In the second segment (sentence 4 above), I didn't denote it very well, but what I used was 60/t to represent the constant velocity during the second segment and time being 90-2t (with t being the time during the first and 3rd segments). The mistake I made was adding 260+120 and getting 400...

Thank you for considering my problem. For the 1st segment, s=1/2at^2, 60=at^2, a=60/t^2. Then, v=at, v=(60/t^2)(t), v=60/t. For the 2nd segment, s=vt, 260=(60/t)(90-2t), 260=5400/t - 120t/t, 260= (5400/t-120), ...Whoops- I just saw my arithmetic error. Got it. Anyway, thank you again for...

I think that the initial acceleration and the deceleration are the same, ditto for the time for phase 1 and 3 so it seems to me that there are really 4 unknowns. I have tried 320= 2(.5a(90-t)^2))+(at)t in various forms but I always wind up with 2 unknowns and I can't find think of another...

Homework Statement A building is 320 meters tall. An elevator accelerates uniformly for 30 meters, travels at a constant rate for 260 meters, then decelerates uniformly for the last 30 meters. The total time is 90 seconds. What is the velocity for the middle 260 meters? Homework...