Why my expression for acceleration doesn't work? So i have an object that moves in a straight line with initial velocity $v_0$ and starting position $x_0$. I can give it constant acceleration $a$ over a fixed time interval $t$. Now what i need is that when the time interval ends this object should stop exactly at a point $x_1$ with it's velocity being equal to $0$. I need to find acceleration $a$ that i can give it in order for that to happen. 
The way i see it we've got a system of equations: 
$$ 0 = v_0 + a t $$
$$ x_1 = x_0 + v_0 t + \frac {a t^2} {2} $$ 
I have only one unknown, which is $a$. 
Let's get $a$ from the first equation: 
$$ a = \frac { - v_0 } { t } $$ 
And put it into the second one: 
$$ x_1 = x_0 + v_0 t + \frac { - v_0 t } {2} $$ 
Now let's express initial velocity ($v_0$) from that equation: 
$$ x_1 - x_0 = v_0 t + \frac { - v_0 t } {2} $$
$$ \frac { x_1 - x_0 } { t } = v_0 + \frac { - v_0 } {2} $$
$$ \frac { 2 ( x_1 - x_0 ) } { t } = 2 v_0 - v_0 $$
$$ v_0 = \frac { 2 ( x_1 - x_0 ) } { t } $$ 
And put it back into equation for acceleration: 
$$ a = \frac { - v_0 } { t } $$
$$ a = \frac { - \frac { 2 ( x_1 - x_0 ) } { t } } { t } $$
$$ a = - \frac { 2 ( x_1 - x_0 ) } { t^2 } $$ 
So we got an acceleration that i need to apply to an object over a time interval $t$, so that it would stop at $x_1$ with velocity $0$, right? 
But it doesn't work! 
Because it doesn't depend on initial velocity at all! So if my object is flying at 2 m/s then i would need to apply the same acceleration as if it was flying 100 m/s, or 1000 m/s? How come? 
Where am i being wrong? This all seems mathematically sound... Am i setting the wrong premises? Interpreting results in the wrong way? 
I really need it for my project, and i've been trying to solve this for weeks, studying different aspects of maths that might help me, but i just can't do it :( 
But this looks so simple! And yet i just can't do it. 11 years of school seem so useless right now... 
Help please 
 A: I will use $t_0$ rather than $t$, since this is also a fixed quantity.
What you are doing doesn't work for arbitrary $t_0$, $x_0$, $x_1$, and $v_0$. 
Since your only unknown is supposed to be $a$, from the first equation you get
$$a = -\frac{v_0}{t_0}$$
From the second equation you get
$$a = \frac{2(x_1-x_0-v_0t_0)}{t_0^2}$$
Thus, for a solution to exist, you must have
$$-\frac{v_0}{t_0} = \frac{2(x_1-x_0-v_0t_0)}{t_0^2}$$
or 
$$v_0t_0 = 2(x_1-x_0).$$
If this does not hold, then there is no solution.
Conversely, if $v_0t_0=2(x_1-x_0)$, then your solution is $a=-\frac{2(x_1-x_0)}{t_0^2} = -\frac{v_0}{t_0}.$
So the solution only exists for a specific value of $v_0$ (given the distance and time), and then the acceleration does depend on the initial velocity.
Alternatively, you can fix any three of $v_0$, $x_0$, $t_0$, and $x_1$, and then solve for the remaining unknown and $a$; but in general you cannot arbitrarily specify all four quantities.
A: As one of the comments suggested, it is not possible for all time intervals $t$. I ignored the fixing of the time interval and solved the problem without that constraint. Hope it helps:
EDIT: If you want to still have a fixed $t_1$, you can view the obtained equation for $t_1$ as a constraint, that must hold, for the problem to have a solution.

Equations of motion of the object:
$$a(t) = a_0$$
$$v(t) = v_0 + a_0t$$
$$x(t) = x_0 + v_0t + \frac{a_0}{2}t^2$$

Equations that have to be satisfied:
$$x_1 = x_0 + v_0 \cdot t_1 + \frac{a_0}{2}t_1^2$$
$$v_1 = v_0 + a_0 \cdot t_1$$
Where $t_1$ is the time, when the object is at $x_1$, so $x(t_1) = x_1$
You have two unknowns: $a_0$ and $t_1$.
Proceeding as you described, you get: 
$$a_0 = -\frac{v_0}{t_1}$$
Plugging this into the first equation, that must be satisfied, gives:
$$x_1 = x_0 + v_0 \cdot t_1 + \frac{-\frac{v_0}{t_1}}{2}t_1^2$$
$$t_1 = \frac{2(x_1-x_0)}{v_0}$$
Plug this into the obtained equation for $a_0$:
$$a_0 = -v_0 \cdot \frac{v_0}{2(x_1-x_0)}$$
$$a_0 = - \frac{v_0^2}{2(x_1-x_0)}$$
A: You are impossing to much things. Think about it this way. If you begin at speed $v_0$ and pretend to have a constant acceleration such that, in a time $t$ you reach $0$ speed, then this acceleration must be:
$$
a =\frac{0-v_0}{t}
$$
There you have the dependency of $a$ with $v_0$, but you won't be able to make it to stop at a point $x_1$, that is, the stopping point would be part of the solution:
$$
x_1 = x_0 + v_0t + \frac{1}{2}at^2
$$
The same way, if you really want to stop at a point $x_1$ in a time $t$, deccelerating at a constant rate, then you get a formula for $a$ that does not depend on $v_0$, as $v_0$ would be part of the solution:
$$
v_0 = 0-at
$$
(Finally you could impose the initial velocity $v_0$ and the stopping point $x_1$, but then the time needed would be part of the solution)
A: (This is too long for a comment, so I'm posting it here.)
There is nothing wrong with your formula:  it is dimensionally correct and your derivation is handled properly algebraically.  The issue that is causing you concern is because the formula obscures the relationship between initial velocity $ \ v_0 \ $ and the initial and final positions $ \ ( \ x_0 \ \ , \ \ x_1 \ ) \ $ and the time interval $ \ t \ $ required to come to rest.
Using your example initial velocites, we'll consider what the quantities are for the same deceleration $ \ a \ = \ -1 \ \frac{\text{m.}}{\text{sec.}^2} \ \ . \ $
$$ \ \ \ v_0 \ = \ 2 \ \frac{\text{m.}}{\text{sec.} } \ \ \Rightarrow \ \ T \ = \ \frac{0 - 2}{-1} \ = \ 2 \ \text{sec.} \ \ \Rightarrow \ \ x_1 - x_0 \ = \ \frac{ 2 \ \frac{\text{m.}}{\text{sec.}} · 2 \ \text{sec.} }{2} \ = \ 2 \ \text{m.} \ \ \ ;  $$
$$ \ \ \ v_0 \ = \ 100 \ \frac{\text{m.}}{\text{sec.} } \ \ \Rightarrow \ \ T \ = \ \frac{0 - 100}{-1} \ = \ 100 \ \text{sec.} $$ $$ \Rightarrow \ \ x_1 - x_0 \ = \ \frac{ 100 \ \frac{\text{m.}}{\text{sec.}} · 100 \ \text{sec.} }{2} \ = \ 5000 \ \text{m.} \ \ (5 \ \text{km.}) \ \ \ ;  $$
$$ \ \ \ v_0 \ = \ 1000 \ \frac{\text{m.}}{\text{sec.} } \ \ \Rightarrow \ \ T \ = \ \frac{0 - 1000}{-1} \ = \ 1000 \ \text{sec.} $$ $$ \Rightarrow \ \ x_1 - x_0 \ = \ \frac{ 1000 \ \frac{\text{m.}}{\text{sec.}} · 1000 \ \text{sec.} }{2} \ = \ 500,000 \ \text{m.} \ \ (500 \ \text{km.}) \ \ \ .  $$
If you had these quantities we've derived as your starting point, your formula produces
$$ a \ \ = \ \ -\frac{2 \ · \ 2 \ \text{m.} }{( 2 \ \text{sec.} )^2} \ \ = \ \ -1 \ \frac{\text{m.}}{\text{sec.}^2} \ \ \ ; $$
$$ a \ \ = \ \ -\frac{2 \ · \ 5000 \ \text{m.} }{( 100 \ \text{sec.} )^2} \ \ = \ \ -1 \ \frac{\text{m.}}{\text{sec.}^2} \ \ \ ; $$
$$ a \ \ = \ \ -\frac{2 \ · \ 500,000 \ \text{m.} }{( 1000 \ \text{sec.} )^2} \ \ = \ \ -1 \ \frac{\text{m.}}{\text{sec.}^2} \ \ \ . $$
The initial velocity of the object coming to rest is imbedded in your formula, so it is "invisible" but it affects the magnitudes of the "stopping distance" and "stopping time" for that object.  The formula does make sense, but you apparently didn't try it out with some measured quantities from a physical situation.
