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?

  • $\begingroup$ You are probably missing that the formula assumes both velocities to be pointing into the same direction. I would guess that for antiparallel velocities the sign changes. $\endgroup$ – CuriousOne May 14 '16 at 20:11
  • $\begingroup$ But even if the sign changes, wouldn't it be the same with the square ? $\endgroup$ – Joker 00 May 14 '16 at 20:13
  • $\begingroup$ $(v_0)^2-(v_1)^2$ is not the same as $(v_0)^2+(v_1)^2$. $\endgroup$ – CuriousOne May 14 '16 at 20:22
  • $\begingroup$ You mean the whole equation would change, not only the velocity's sign ? $\endgroup$ – Joker 00 May 14 '16 at 20:23
  • 1
    $\begingroup$ If the equation is correct (and I didn't think for a second about that because it doesn't interest me), then my best guess is that the real equation has a slightly different form that can be reduced to this one with the sign change. Did you try to calculate it, yourself? Where did you get this from? $\endgroup$ – CuriousOne May 14 '16 at 20:25

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.


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.

$$a \le \dfrac{0^2 - (v_0-v_1)^2}{2d}$$ $$a \le -\dfrac{(v_0-v_1)^2}{2d}$$


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
$a =(i^2-u^2)/2d$ 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!

  • $\begingroup$ This equation makes more sense to me but I want to understand the first one. It's different right ? $\endgroup$ – Joker 00 May 14 '16 at 20:19
  • $\begingroup$ hope you understand after the edit! $\endgroup$ – Ayush Jain May 14 '16 at 20:22
  • $\begingroup$ But what if they are moving in the opposite directions, the result will be equal to 0 and that's not correct, is it? $\endgroup$ – Joker 00 May 14 '16 at 20:29
  • $\begingroup$ the first part of my answer explains when they move in opposite direction i already have given you the result and that can't be zero! The result in your question is for the cars when they are moving in same direction! $\endgroup$ – Ayush Jain May 14 '16 at 20:32

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