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Mechanics Question [2]

  1. Sep 5, 2012 #1
    1. The problem statement, all variables and given/known data

    A building have 198 meters, an elevator take 40 seconds to go all the way up, knowing that the elevator takes 6 secods to speed up and another 6 seconds to speed down and the aceleration is constant during this period. Determine the top speed of the elevator.


    2. Relevant equations

    (1) R = Ro + Vot +at²/2
    (2) V² = Vo² + 2aΔt


    3. The attempt at a solution

    I tryed to R1 + R2 + R3 = 198

    Now using (1) in R1 in this aproach Ro is always zero, and in R1 Vo is also zero
    R1 = 18a

    For R2 we have aceleration zero but we have initial speed (the final speed of R1) so we have
    R2 = 28V1

    and R3 = R1 since same aceleration only in oposite direction.
    R3 = 18a

    for R2 I elaborated V1 using (2)

    V1² = Vo² + 2aR1
    V1² = 0 + 36a²
    V1 = 6a

    use that on the R2 formula:

    R2 = 28xV1
    R2 = 168a

    Now going to the final formula:

    I tryed to R1 + R2 + R3 = 198

    18a + 168a + 18a = 198
    a = 0.9705

    now using the a on the first 6 secs of the aceleration.

    V = at
    V = 0.9705*6
    V = 5,823529 m/s

    The list is telling me the correct answer should be 6.... where did I go wrong? The list answer may be wrong? Do you guys have another aproach for this exercise? I would like to see it.

    Thank you...
     
    Last edited by a moderator: Sep 5, 2012
  2. jcsd
  3. Sep 5, 2012 #2

    vela

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    You work looks fine, [strike]but your calculated value for a is off. I got a=0.99 m/s2[/strike]. Keep in mind sig figs as well.

    EDIT: Never mind. I can't add.
     
    Last edited: Sep 5, 2012
  4. Sep 5, 2012 #3
    I got the same answer too.

    34v=198
    v=5.82352
     
  5. Sep 5, 2012 #4
    How you got to 34v = 198??
     
  6. Sep 5, 2012 #5
    It can be easily shown if you sketch the velocity vs time graph.

    vt=s
    vx6x1/2 +vx6x1/2 + 28xv= 198
    34v=198
     
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