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Solving for time given a messy displacement equation

  1. Dec 29, 2015 #1
    I'm having some trouble with this problem (adapted from K&K 1.23).
    1. The problem statement, all variables and given/known data
    An elevator is programmed to start from rest and accelerate according to
    \begin{align*}
    a(t) &= \frac{a_m}{2}\left[1 - \cos\left(\frac{2\pi t}{T}\right)\right] &\mbox{for }0\leq t\leq T \\
    a(t) &= -\frac{a_m}{2}\left[1 - \cos\left(\frac{2\pi t}{T}\right)\right] &\mbox{for }T\leq t\leq 2T
    \end{align*} where ##a_m## is the maximum acceleration and ##2T## is the total time for the trip.
    What is the time required for a trip of distance ##D##?

    2. Relevant equations
    ##v = \int adt##
    ##x = \int vdt##

    3. The attempt at a solution
    Assume for a moment that ##0 \leq t \leq T##. Integrating then yields
    \begin{align*}
    x &= \frac{a_m}{2}\left[\frac{t^2}{2} + \left(\frac{T}{2\pi}\right)^2\cos\frac{2\pi t}{T}\right]
    \end{align*}
    Letting ##t = 0##, we find
    \begin{align*}
    x_0 &= \frac{a_m}{2}\left(\frac{T}{2\pi}\right)^2
    \end{align*}
    It follows that
    \begin{align*}
    D &= x - x_0 = \frac{a_m}{2}\left[\frac{t^2}{2} + \left(\frac{T}{2\pi}\right)^2\left(\cos\frac{2\pi t}{T} - 1\right)\right]
    \end{align*}
    But I'm not seeing a way to hammer this into a useful expression for ##t##. Perhaps there is an error in the way the problem is worded?
     
  2. jcsd
  3. Dec 29, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    The time for the half trip is T. Have you tried substituting T for t in your distance equation?
     
  4. Dec 29, 2015 #3
    I'm not sure I understand. How would doing that help me find a general expression for ##t##, given any value of ##D##?
     
  5. Dec 29, 2015 #4
    If I've interpreted the question correctly, D is the distance for the whole trip, which takes a total time 2T, so . . . . . (gneill is pointing you in the right direction)
     
  6. Dec 29, 2015 #5
    Ah, I thought ##D## was any point along the trip. Sorry for the confusion.
     
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