Car Crash Scenario Two vehicles travelling at 80mph in the same direction.
Vehicles are directly behind each other.
12 meter distance between them.
Front door of car in front rips of and hits the front window of the car behind.
At what speed did the door make contact with the car?
How would one go about calculating this?
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My current guess is as follows: The car behind would take approximately 0.34 seconds to reach the door in front, the door in front would have a forward velocity starting at 80mph but would quickly be decelerating due to no longer being attached to the car in front. Considering the car behind reach the door in approximately 0.34 seconds, i would guess that the speed of the door would be still very close to 80mph.
But this is a pure guess, is there a formula i can use here to accurately calculate this?
 A: The door is not slowing down because of its mere detachment from the front car. It is slowing down because of air drag on the door.  That drag would depend on the area of cross-section interaction, the mass of the door and whether the door is tumbling.  One might be able to estimate a worst-case scenario, but there isn't really a way to calculate an accurate.
A: Once the door is detached from the front car, it will start decelerating, because of air drag, friction with the floor, etc. You can start assuming that this decelaration is constant, it is not perfectly accurate, as the door will likely be rotating, and being an irregular form, the drag will vary chaotically. Also the drag vary with velocity.
If you assume this door (de-)acceleration constant, then the calculation is easy, particularly, if you do the computation from the frame of reference of the moving cars (lowercase relativity!). The distance travelled by the door is:
$$    D_d = D_a * t^2 $$
being, $D_d$ the distance, $D_a$ the door acceleration (to be determined experimentally) and $t$ the time.
Now the rest is easy: the crash will be when $D_d = 12$.

You ask how to calculate this accurately, but that is impossible. Even if you were to do the actual experiment (please use a wind tunnel, not real roads!), you would get quite different results from one execution to the next, simply because the movement of the door in the air will be chaotic and unpredictable.
A: If the door has only air drag acting on it, its acceleration is $a=-\beta v^2$ where $\beta$ is some drag coefficient and $v$ is speed. The time and distance needed to slow down from $v_1$ to $v_2$ is defined by
$$ \begin{aligned} 
  \Delta t & = \int \limits_{v_1}^{v_2} \frac{1}{a}\,\mathrm{d}v = \frac{1}{\beta} \left( \frac{1}{v_2} - \frac{1}{v_1} \right) \\
  \Delta x & = \int \limits_{v_1}^{v_2} \frac{v}{a}\,\mathrm{d}v = -\frac{1}{\beta} \ln \left( \frac{v_2}{v_1} \right)
\end{aligned}  $$
Combine the above to get the kinematics of the flying door
$$ \Delta x = \frac{1}{\beta} \ln \left(1+\beta v_1 \Delta t \right) $$
In the same interval the car in the back travels the distance $$\Delta x = v_1 \Delta t - s$$ where $s$ is the initial separation (15 m)
Unfortunately the equation  $ \frac{1}{\beta} \ln \left(1+\beta v_1 \Delta t \right) = \Delta t v_1 - s$ cannot be solved for $\Delta t$ (the flight time) analytically. A numerical method is needed. Once $\Delta t$ is known the absolute speed of the door is
$$ v_2 = \frac{v_1}{1+\beta v_1 \Delta t} $$
Finally the impact speed is $v_{imp} = v_1 - v_2$.
NOTES: The deceleration of the door will vary from the equation stated because as the door rotates in the air the frontal area exposed to air drag changes. An average will have to be assumed.
A: There's a good way to get an estimate out of this.
First, perform dimensional analysis to get an estimate.
Start with, finding ${{\Delta a} \over {\Delta v}}$, which is the change in the drag acceleration with respect to velocity. The argument that follows is physical rather than technical, so beware. The above expression, should be proportional to the density of air $\rho$ and the surface area exposed $A$. In addition, the more mass the object has, the greater its momentum, and the harder it is to accelerate. Thus, ${{\Delta a} \over {\Delta v}}$ is inversely proportional to mass. Finally, the faster an object changes its velocity the greater the change in force. So the above is proportional to $v$.
Putting the above together yields,
$${{d a} \over {d v}}=C_d \cdot {{\rho \cdot A} \over {m}} \cdot v$$
Where we have changed $\Delta$ to $d$ since we'll examine this at infinitesimal changes in the variables. $C_d$ is referred to as the "drag coefficient". Summing the above expression over all changes in v, will yield the drag acceleration.
$$da=C_d \cdot {{\rho \cdot A} \over {m}} \cdot v \ dv$$
$$\rightarrow a_d=C_d \cdot {{\rho \cdot A} \over {2 \cdot m}} \cdot v^2$$
We have to make a couple more assumptions now, and then we can get the total acceleration on the door. First of all, when the door detaches from the car, we assume it'll hit the window on the wide side of the door. Thus, we will simplify the door to a rectangular block. Secondly, we'll assume the door does not spin while in the air. Third, we assume that the speed is fast enough to ignore Reynolds number phenomena. Last, we will assume that door movement is purely horizontal. Using these assumptions, we get,
$$a=C_d \cdot {{\rho \cdot A} \over {2 \cdot m}} \cdot v^2=\lambda \cdot v^2$$
Where $\lambda=C_d \cdot {{\rho \cdot A} \over {2 \cdot m}}$.
To solve this equation, we notice that the rate of change of velocity is acceleration.
$${{dv} \over {dt}}=\lambda \cdot v^2$$
$$\rightarrow {{dv} \over {\lambda \cdot v^2}}=dt$$
We once again sum over the $dv$ and $dt$ and get, taking the integral.
$${{-1} \over {\lambda \cdot v}}=t$$
$$\rightarrow v={{-1} \over {\lambda \cdot t}}$$
Of course, now we note that we have an initial velocity term and proceed to add it in the denominator so that at $t=0$ the velocity is 80mph or $35.75$ meters per second.
$$\rightarrow v={{1} \over {\lambda \cdot t+1/v_0}}$$
Where we've eliminated a sign, and have created $v_0$ for the original velocity.
Note that velocity is the rate of change of position, and integrate again.
We get,
$$x={1 \over {\lambda}} \cdot \ln(\lambda \cdot v_0 \cdot t+1)$$
Sadly, after all this work, you can't solve the equation to get the time to impact the car.
The drag coefficient for a plate is $2$. The density of air is $1.2754$ kilograms per $m^3$. The surface area shall be estimated as 1 $m^2$. The weight will be $30$ kilograms. The initial velocity is $35.75$ meters per second.
Substituting this with the condition for collision gives.
$$v_0 \cdot t-12={1 \over {\lambda}} \cdot \ln(\lambda \cdot v_0 \cdot t+1)$$
Solving yields the time of collision at $t=0.836$. Hope you like overkill ;)
Here's a reference to the concept of free fall, and the forces that act on objects in mid fall.
Conclusion
In other words, this envelope calculation basically shows that the problem is ill-posed. Since the door is to hit the road before the car, there can be no scenario where it hits a window. At this point, the reader should accept this is just a home-work problem and move on...
(Posted from mobile)
