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Homework Help: Acceleration, constant velocity, deceleration

  1. Sep 14, 2013 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations
    s=.5at^2, v=at, s=vt

    3. The attempt at a solution I tried 320m:290m=v(final)sec:3.5556 sec. This yields a ball park answer but I don't think it's right. I'd be very grateful for any help.
  2. jcsd
  3. Sep 14, 2013 #2
    You have six unknowns: initial acceleration; constant velocity; final deceleration; the duration of the initial acceleration; the duration of the constant velocity motion; the duration of the final deceleration.

    So you need six equations to find all the unknowns.

    The problem directly describes, in words, four equations. You need to figure out what two additional equations and solve the system.

    You may find a shortcut if you think about the physical significance of the fact that the lengths of the initial and final segments are equal, but that is not strictly necessary.
  4. Sep 14, 2013 #3
    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 equation. Am I barking up the wrong tree?
  5. Sep 14, 2013 #4
    You are right, the initial acceleration and the final deceleration are the same in magnitude, so only four unknowns.

    There are: ##a##, the acceleration; ##v##, the constant velocity; ##t_1##, the duration of acceleration; ##t_2##, the duration of the uniform motion.

    What equations do you have?
  6. Sep 15, 2013 #5
    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 your very kind assistance.
  7. Sep 15, 2013 #6
    It seem to me that you are using the same unknown "t" for both the first and the second segment.

    Is this what you have spotted? Have you solved the problem?
  8. Sep 15, 2013 #7
    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. That will yield a believable answer for the constant velocity but won't check when you use it to confirm the distance of the first segment that would be produced by the derived acceleration and the calculated time of 14.2 seconds. Adding 260 + 120 and getting 380 makes everything check. In any case, I am grateful to you for taking the time to consider my question.
  9. Sep 15, 2013 #8
    Note that the problem is symmetric: equal time/duration for acceleration and deceleration. Hence you can reduce this to a "half-problem": the elevator accelerates for 30 meters, then rides uniformly for 130 meters, 45 seconds total. I think this should be much easier to solve.
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