Solving for time given a messy displacement equation

In summary: So if ##D## is the total distance of the trip, then the time required would be ##2T##. Is that correct?In summary, the conversation was about a problem involving an elevator with a specific acceleration function, and the task was to find the time required for a trip of a given distance. The solution involved integrating the acceleration function and substituting the given values to find a general expression for time. However, there was some confusion about the given distance and it was eventually determined that the time required for the trip would be 2T.
  • #1
Rations
5
0
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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  • #2
The time for the half trip is T. Have you tried substituting T for t in your distance equation?
 
  • #3
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##?
 
  • #4
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)
 
  • #5
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.
 

1. How do I solve for time given a messy displacement equation?

First, identify the variables present in the equation (e.g. displacement, velocity, acceleration). Then, use the given values and equations of motion to rearrange the equation and isolate the variable for time.

2. What are the steps to solving for time in a displacement equation?

The steps may vary depending on the specific equation, but generally, you will need to rearrange the equation to solve for time. This may involve using algebraic manipulation, substitution, or plugging in values from other equations.

3. Can I use any equation to solve for time in a displacement problem?

Yes, as long as you have enough information to solve for time. However, some equations may be more suitable for certain types of problems (e.g. constant acceleration vs. variable acceleration).

4. What units should my time answer be in when solving for time?

The units for time will depend on the units of the other variables in the equation. For example, if the displacement is given in meters and the velocity is given in meters per second, the time answer will be in seconds.

5. How can I check if my solution for time is correct?

You can plug your answer for time back into the original equation and see if it satisfies the equation. Additionally, you can use common sense and reasoning to make sure your answer is reasonable and makes sense in the context of the problem.

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