# Homework Help: Solving for time given a messy displacement equation

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1. Dec 29, 2015

### Rations

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. Dec 29, 2015

### Staff: Mentor

The time for the half trip is T. Have you tried substituting T for t in your distance equation?

3. Dec 29, 2015

### Rations

I'm not sure I understand. How would doing that help me find a general expression for $t$, given any value of $D$?

4. Dec 29, 2015

### TonyS

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)

5. Dec 29, 2015

### Rations

Ah, I thought $D$ was any point along the trip. Sorry for the confusion.