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Homework Help: Constant Acceleration

  1. Sep 22, 2010 #1
    Going Up An elevator cab in a New York hotel has a total run of 185 m. Its maximum speed is 261 m/min. Its acceleration (both speeding up and slowing down) has a magnitude of 1.21 m/s2.

    (1) How far does the cab move while accelerating to full speed from rest?

    v = 261m/min
    = 261m/min * (1/60)min/s
    = 261/60 m/s
    = 4.35m/s
    a = 1.21m/s^2

    a) u = 0
    v^2 = u^2 + 2ad
    v^2 = 2ad
    d = v^2 /2a ≈ 7.8192m

    (2) How long does it take to make the nonstop 185 m run, starting and ending at rest?

    so d = 2*7.8192 = 15.6384m all together
    d = (1/2)a(t1)^2
    t1 = √(2d/a) ≈ 5.08414s
    spends D = 185 - 15.6384 = 169.3616m at constant velocity v=4.35m/s
    D = vt2
    t2 = D/v ≈ 38.9337s
    T = 5.08414 + 38.9337 = 44.01784s all up

    The issue i'am having is with question 2. From some reason 44.01784 isn't right, what am i doing wrong. i know my first answer is correct
  2. jcsd
  3. Sep 22, 2010 #2
    Your mistake is that, you have to calculate the time separately for the two accelerations.
    what you did was take the whole distance of acceleration (the one at the beginning and at the end) and calculated time for that. Which means that you actually over calculated the time.
    find the time it takes for speeding up, and multiply it by 2. that will be the total acceleration time.
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