Solving for time given a messy displacement equation

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I'm having some trouble with this problem (adapted from K&K 1.23).

Homework Statement


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##?

Homework Equations


##v = \int adt##
##x = \int vdt##

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?
 
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The time for the half trip is T. Have you tried substituting T for t in your distance equation?
 
gneill said:
The time for the half trip is T. Have you tried substituting T for t in your distance equation?
I'm not sure I understand. How would doing that help me find a general expression for ##t##, given any value of ##D##?
 
Rations said:
I'm not sure I understand. How would doing that help me find a general expression for ##t##, given any value of ##D##?
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)
 
Rations said:
I'm not sure I understand. How would doing that help me find a general expression for ##t##, given any value of ##D##?
TonyS said:
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)
Ah, I thought ##D## was any point along the trip. Sorry for the confusion.