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 

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.
