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!!!
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.