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sorry for disturbing your time. But I just dont know how to solve this problem.. Its about rear end collisions.

Let's assume the ego vehicle is driving with 70 km/h and the car behind us aswell. Now we have to decelerate because there is an imminent collision infront.

The thing is that the deceleration must be limited so that the severity of the rear-end collision is not too high when the vehicle behind hits us..

So for example both cars drive with 70 km/h at the initial distance of 17,5m. We start decelerating with a not given value. From the moment the brake lights start to come on, the car behind us starts to brake with a constant deceleration of 6 m/s^2 after 1.8s. So the reaction time is 1.8s.

Now to the question: What is the necessary deceleration from the ego vehicle, so that at the time of the collision, the differential speed is less than 10 km/h.

The problem is that there is no time given when the collision will occur and therefore there are 2 unknown parameters..

Please help me!!!

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  • $\begingroup$ Do you assume the vehicle behind decelerates at the same rate as yours? $\endgroup$
    – Karl
    Jun 27 at 17:20
  • $\begingroup$ The collision time depends on the deceleration rate, so you can solve for that first and then evaluate the speed difference. $\endgroup$
    – Karl
    Jun 27 at 17:21
  • $\begingroup$ Sorry I forgot to add.. The vehicle behind me starts decelerating with 6 m/s^2 after the reaction time of 1.8s. The deceleration rate of the ego car is not given, which needs to be calculated by a function so that at the time of the collision, the differential speed is not less than 10 km/h $\endgroup$
    – Matahari
    Jun 27 at 17:27
  • $\begingroup$ Have you tried calculating the collision time? $\endgroup$
    – Karl
    Jun 27 at 17:54
  • $\begingroup$ What is an "ego" vehicle ? What is the initial distance between it and the vehicle behind ? $\endgroup$ Jun 27 at 18:17

1 Answer 1

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You have two differential equations, one for each vehicle:

$$ \begin{align} \ddot{x}_1(t) &= a_1 \\ \ddot{x}_2(t) &= a_2 \end{align} $$

with positions $x_1,x_2$, velocities $\dot{x}_1,\dot{x}_2$ and constant deceleration inputs $a_1,a_2$. The differential equations have the solution for velocity:

$$ \begin{align} \dot{x}_1(t) &= \dot{x}_1(0) + a_1 t \\ \dot{x}_2(t) &= \dot{x}_2(0) + a_2 t \end{align} $$

and for position:

$$ \begin{align} x_1(t) &= x_1(0) + t \dot{x}_1(0) + \frac{1}{2}t^2 a_1 \\ x_2(t) &= x_2(0) + t \dot{x}_2(0) + \frac{1}{2}t^2 a_2 \end{align} $$

You have $x_1(0) = 17.5, x_2(0) = 0, \dot{x}_1(0) = \dot{x}_2(0) = \frac{175}{9}$, $a_1$ constant and unknown and

$$ a_2=\begin{cases}0 & t < 1.8\\-6 & t \geq 1.8\end{cases} $$

We search for $a_1 < 0$ such that $\dot{x}_1(t_c) - \dot{x}_2(t_c) = \frac{25}{9}$ where $t_c$ is the unknown collision time such that $x_1(t_c) = x_2(t_c)$.

In the first $1.8$ seconds the second vehicle has $a_2 = 0$ so

$$ \begin{align} x_1(1.8) &= \frac{81}{50} a_1 + \frac{105}{2} \\ x_2(1.8) &= 35 \\ \dot{x}_1(1.8) &= \frac{9}{5} a_1 + \frac{175}{9}\\ \dot{x}_2(1.8) &= \frac{175}{9} \end{align} $$

The collision time demands

$$ x_1(1.8) + (t_c - 1.8) \dot{x}_1(1.8) + \frac{1}{2}(t_c - 1.8)^2 a_1 = x_2(1.8) + (t_c - 1.8) \dot{x}_2(1.8) + \frac{1}{2}(t_c - 1.8)^2 a_2 $$

So, ignoring the second collision time:

$$ t_c = \frac{54 - \sqrt{-1361 a_1 - 5250}}{5(a_1 + 6)} $$

Now consider $\dot{x}_1(t_c) - \dot{x}_2(t_c) = \pm\frac{25}{9}$ so

$$ \frac{9}{5}a_1 + 6(t_c - 1.8) + a_1(t_c - 1.8) = \pm\frac{25}{9} $$

The first collision happens when $\dot{x}_1(t_c) - \dot{x}_2(t_c) = -\frac{25}{9}$ with

$$ a_1 = -\frac{440875}{110241} \approx -3.9992 $$

For 20 km/h you just take $\dot{x}_1(t_c) - \dot{x}_2(t_c) = -\frac{50}{9}$ instead of $\dot{x}_1(t_c) - \dot{x}_2(t_c) = -\frac{25}{9}$. Then you get a different $a_1$ namely

$$ a_1 = -\frac{487750}{110241} \approx -4.4244 $$

and so on.

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  • $\begingroup$ Thank you very much for the solution! Still I have some questions.. What do you mean with "ignoring the second collision time", and which a2 value do you use there? The -6 deceleration value? $\endgroup$
    – Matahari
    Jun 28 at 6:35
  • $\begingroup$ And if I would first calculate which ego speed i have after 1.8s and then use this speed with the calculated a1 from now on for the following equations: equation: x1=x2 (to get the collision time) and insert this collision time in x'1(tc)=x'2(tc) I get a differencial speed of -1.05 m/s. So no collision would happen $\endgroup$
    – Matahari
    Jun 28 at 6:54
  • $\begingroup$ Please help me.. $\endgroup$
    – Matahari
    Jun 28 at 12:36
  • $\begingroup$ @Matahari I use the $a_2$ value stated in the answer... Did you try to follow (and understand) each step in the calculation? If not, you should first try to do that before hastily using $a_1$, as it is only valid under the assumptions stated in your question. With second collision time I mean the second solution to the quadratic equation. $\endgroup$
    – SampleTime
    Jun 28 at 15:46
  • $\begingroup$ I guess I understood your calculation.. Because you want to see the situation with (t-1.8s) after the reaction time, so when the car behind us starts to decelerate. I have realized that the deceleration value is valid, but I have then tried for another desired valicity difference, f.e. 20 km/h. The deceleration I then got, was not working.... Could you possibly explain in some few words or also with an example for example how you got to the term tc, because I have tried to also calculate with the same values, but didnt reach to the same solution as yours.. $\endgroup$
    – Matahari
    Jun 28 at 17:14

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