I don't understand the formula of constant deceleration required to prevent a vehicle collision As I was looking for the formula that an Intelligent Speed Assist technology use to calculate the required deceleration that a vehicle must apply avoid collision, I found this one: 
$$a = \frac{V^2_0 - V^2_1}{2d}$$
I think I don't understand it well because for me, if $V_0$ (which is the initial velocity of the host vehicle ) is equal to the opposite vehicle's velocity $V_1$ and that they are moving to opposite direction, the formula would return a required deceleration of zero which is far from being correct.
Can someone help me see what I am missing? 
 A: I'm getting a different formula. Please someone point out my error if you see it.
So the host vehicle is to the left of the target vehicle. In order to avoid collision, in the worst case scenario, by the time the host vehicle decelerates to the target vehicle's velocity, the target vehicle is still to the right of the host vehicle. We're assuming positive is to the right, negative is to the left, and $d>0$.
Assume t is the time till the velocities become equal. Let $d_0$ be the position of the host vehicle (that accelerates). Let $d_1$ be the position of the other vehicle. Assume the host vehicle is at 0 position at t=0, and opposite vehicle is at position d at $t=0$. The final velocity of the host vehicle is $v_1$. We assume $v_1<v_0$, otherwise there's no danger of collision. So $a<0$.
$$a = \dfrac{v_1-v_0}{t}$$
$$t = \dfrac{v_1-v_0}{a}$$
$$d_0 = v_0\times t + \dfrac{1}{2}at^2$$
$$d_1 = v_1\times t + d$$
We require $d_1 \ge d_0$ for no collision. Here, $v_1,v_0$ may be positive or negative. 
$$v_1 \times t + d \ge v_0 \times t + \dfrac{1}{2}at^2$$
substitute in the formula for t
$$v_1 \times (\dfrac{v_1-v_0}{a})+d \ge v_0 \times \dfrac{v_1-v_0}{a}+\dfrac{1}{2}a(\dfrac{v_1-v_0}{a})^2$$
$$\dfrac{(v_1-v_0)^2}{2a}+d \ge 0$$
$$d \ge -\dfrac{(v_1-v_0)^2}{2a}$$
$$a \le -\dfrac{(v_1-v_0)^2}{2d}$$ (since a is negative the greater than becomes less than)
This makes more sense to me since it accounts for equal speeds but opposite velocities for the host and other vehicles.
EDIT:
Much simpler solution. Go to the rest frame of the target vehicle. Initial velocity of host vehicle is $v_0-v_1$, final velocity is 0. Distance between two cars is still d. Just use the standard kinematics formula with a,d and initial and final velocities.
$$a \le \dfrac{0^2 - (v_0-v_1)^2}{2d}$$
$$a \le -\dfrac{(v_0-v_1)^2}{2d}$$
A: if we assume the initial velocity of host car to be u
and initial velocity of other car as i
relative velocity of other car with respect to te first one would be (u+i).
Let initial separation be d
applying equation of kinematics
$v^2 = u^2 + 2as$
$0 = (u+i)^2 + 2ad$
so,
$a=-(u+i)^2/2d$
Now, required deceleration can be calculated!
Moreover, if the same scenario was with vehicles moving in same direction(forward vehicle moving slower the tailing one), it is to be noted that collision would stop even if their velocities become equal to the velocity of the leading one, then required deceleration can be calculated as
$v^2=u^2+2as$
$i^2=u^2+2ad$
$a =(i^2-u^2)/2d$
and now in this case if speeds of both are equal in same direction collision won't occur, so required deceleration would be 0 and that is correct!
