Calculating Speed of T-Bone Car Accident

[TJU]

Hello,

I am not a big math guy, although I do have a bachelors degree I don't remember any of my formulas from high school lol.

I was involved in a car accident and I am researching the physics of the accident.

I want to try and approximate the speed of the vehicle that hit me broad side.

The car pushed me aprox. 150 to 200 feet, and it was a t-bone accident.

there is car A (theres) and car B (mine)

car A weighed 5,437 pounds and car B weighed 2,654 pounds how fast was car A traveling to have pushed me 150-200 feet. I approximate based on witnesses 50-60 mph but i would like the physics equation and answer if at all possible

thanks and god bless

 

[marcusl]

Hope you are ok. Car accidents are among the hardest problems to calculate because every real-life non-ideal exception to the equations is present. Here are a few of the biggest:

1. Auto collisions are inelastic because considerable energy is dissipating into crumpling and distorting the metal of both cars. Car B probably looks (from a top view) more like a C now than an I, and car A is shorter (i) than when it started (I). Think about how much of A's incoming kinetic energy was dissipated as work to bend the metal into those shapes. Kinetic energy is not conserved in this type of collision, so you need to know this work to solve the problem.

2. Tires don't roll sideways, so energy was dissipated (as friction, heat, and in dissociating rubber molecules) in leaving skid marks as your car was pushed sideways. This is what slowed your car after the collision. You need to know this energy also.

3. Did the cars stick together and move as one? Probably they did for awhile, but maybe not for the entire distance that B moved. If driver A was applying brakes while tangled up, all is more complicated. These matter, and are hard to treat.

In short, you are unlikely to come up with even a rough estimate that is very useful. If you'd like to try, here are the relevant equations:

Conservation of energy

$$\frac{1}{2}m_A v_{A,i}^2+\frac{1}{2}m_B v_{B,i}^2=\frac{1}{2}m_A v_{A,f}^2+\frac{1}{2}m_B v_{B,f}^2+W$$

where i,f indicate initial and final, and W is the work expended during the collision. v_B,i = 0 if your car was stopped at the time of collision. Conservation of momentum is

$$m_A v_{A,i}+m_B v_{B,i}=m_A v_{A,f}+m_B v_{B,f}$$.

Using these equations will give you the instantaneous speed of B immediately after contact.
Now you need to describe how B's speed immediately after collision is scrubbed off to final resting point.

$$\frac{1}{2}m_A v_{A,f}^2=\int \vec{F} \cdot d\vec{s}+W_2$$

where F is the force of friction and W_2 is the work expended in the other things mentioned earlier. You might take F as constant and ignore W_2, as a crude approximation, in which case the right hand side becomes Fs, where s is 150 to 200 feet. In fact, you can take this as a starting point and work backwards to solve for what you want (v_A,i).

If the cars locked together and traveled as one for any time, the above equations get even worse.

As a final remark, you can see from the equations that it is disadvantageous to you to have m_B smaller than m_A. Safety considerations suggest that you want to be in a big heavy vehicle if you are going to be hit by the same. I really hope you came out of this ok.