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+----------------------------->
o                  x1

(X axis is distance and Y is acceleration)

Excuse my crappy ASCII graph but I can't post images so please imagine that the line made of + is actually a concave curve

My question is: knowing x1 how would I calculate the acceleration increase needed to get from O to x1 without overpassing x1?

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It'll help if you could post a fuller explanation, as well as explain more clearly what your problem is, currently the question is very confusing. What kind of a system are you talking about? One can go between two points in space at a constant velocity, without any acceleration as well. Also, what does "overpassing" mean here? –  Kitchi Feb 13 '13 at 11:38
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Do you mean you want to stop at x1? Or reach a particular acceleration at the same time you reach x1? –  Michael Brown Feb 13 '13 at 12:27
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If you have a clearer image please give a link to it and someone with the rep. can edit it into your post. –  Michael Brown Feb 13 '13 at 12:28
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This reads like an XY problem. Please can you add more detail about the broader picture? That is, why do you want this question answered? What will you do with the solution? If you can answer that, we'll be able to give you better answers. –  EnergyNumbers Feb 13 '13 at 12:38
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There are many solutions to this problem depending on your requirements. Do you know the velocity at x1 and x2? Do you have to get from x1 to x2 in a certain amount of time? Do you need to stop at x2? Do you need realistic physics? - If it's a game or something you could maybe get away with just moving at a uniform velocity and stopping on a dime. –  Michael Brown Feb 13 '13 at 12:59

2 Answers 2

up vote -1 down vote accepted

This solution is based on the following interpretation of the OP's question (see the comments beneath the question):

Give a function $a(x)$ such that $a(x)$ is the acceleration of a particle at $x$:

$$ \ddot{x} = a(x) $$

and the motion is subject to the following boundary conditions:

$$ \begin{array}{lcl} x(0) &=& x_1 \\ \dot{x}(0) &=& 0 \\ x(T) &=& x_2 \\ \dot{x}(T) &=& 0 \end{array} $$

where the final time $T$ is unspecified and we take $x_2 > x_1$ for simplicity.

This is an ill-posed problem in the mathematical sense in that there are clearly many solutions. Any function $a(x)$ which obeys the condition

$$ a(\frac{x_1+x_2}{2} - x) = -a(\frac{x_1+x_2}{2} + x) $$

will satisfy the final constraint by symmetry.

A simple solution is

$$ a(x) = A \frac{x_1 + x_2 - 2 x}{x_2 - x_1} $$

for $A>0$. Integrating the equation of motion subject to the initial conditions gives

$$ x(t) = \frac{x_1 + x_2}{2} - \frac{x_2 - x_1}{2} \cos\left( \omega t \right) $$

where

$$ \omega \equiv \sqrt{\frac{2 A}{x_2 - x_1}} $$

The final time $T$, originally unspecified, can be found from the final condition $x(T)=x_2$:

$$ T = \frac{\pi}{\omega} = \pi \sqrt{\frac{x_2 - x_1}{2 A}}$$

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This is helpful, thank you! 1 question though: who is A exactly? –  user1233963 Feb 13 '13 at 14:37
    
$A$ is the maximum acceleration. You can choose it to be whatever you want and it affects the transit time according to the last equation. –  Michael Brown Feb 13 '13 at 14:40
    
Why the downvote? –  Michael Brown Aug 12 '13 at 1:05

If you know the initial velocity and final velocity then you can calculate the acceleration from the $O$ to $X_1$.

$v^2 = u^2 + 2as$

$v = $ velocity at point $X_1$
$u = $ velocity at point $O$
$s = $ distance traveled (which is $X_1$ units)
$a = $ acceleration of the point from $O$ to $X_1$

Here $a$ is nothing but the increment in the acceleration.

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