0
^
|
|
|
|                   +
|                  +|
|                +  |
|              +    |
|            +      |
|          +        |
|        +          |
|      +            |
|    +              |
|  +                |
|+                  |
+----------------------------->
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?

user1233963
  • 103

2 Answers2

0

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.

Waffle's Crazy Peanut
  • 9,178
mathman
  • 1
-1

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 $x(T)=x_2$:

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

Michael
  • 16,830