The Traffic Accident Reconstruction Origin -ARnews-
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The Traffic Accident Reconstruction Origin -ARnews-
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For example, to continue this discussion look for a thread titled
Paint's drag factor
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Our equation |
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Subtract |
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Factor out |
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Find the numbers that will make the left side of the equation equal 0. |
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Thus, the solutions to 
are 0 and 2. Check the
solutions: substitute 2 for x in the original equation and see if it satisfies the equation. Do the same for 0. Notice that a second-degree equation
gives us two solutions. How about 
? The square root method will not work. We can factor this equation, though, into 
. This is done by finding two numbers whose product
is the third term (15), and whose sum is the coefficient of the middle term (8). The two numbers are 3 and 5. Once factored, it is easy to see our solution set is -3 and -5. Check this one too: substitute -3 into the original equation and see if it
satisfies the equation. Do the same for -5.
So far, we have looked at the easy ones. How about an equation such as 
? We cannot take the square root of both sides, nor can we factor. We
can, however, use the quadratic equation.
The quadratic equation is: 
.
To use the quadratic equation, we must write our second-degree equation in standard form; that is, from
the second power, to the zero power. The standard form is written
. Since 
, and 
, we can simplify this to 
. Let's use our previous example,
, to demonstrate the quadratic equation. First, we see that the equation is already in standard form. Therefore, our A, B and C coefficients are 1, 1 and -5, respectively. Substitute these numbers into the quadratic equation:
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Substitute the coefficients into the quadratic equation. |
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Complete the exponent and multiplications |
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This is the answer (it is acceptable to leave the answer in this form because |
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Although factoring and the square root method are the easiest to use, they do not work for all second-degree equations. The quadratic equation will work in all cases, so it is best to know this method.
Systems of Equations
Let's look at a situation where a system of equations may be applied. You are asked to calculate the velocity of two cars given the following information:
One equation will not solve this problem for us. The most obvious equation that we can apply here comes from information that vehicle 1 is traveling 40 mph faster than vehicle 2. To show this algebraically, we write the equation
(Equation A). We can see that there are an infinite number of solutions to this equation: 80=40+40, 70=30+40, 100=60+40, etc. We obviously need more information to solve this problem; this information is provided
by the first two clues. Recall that distance = velocity X time, or 
. When we picture in our minds the two cars, we know that they will meet after 4 hours. We also know that because vehicle 1 is traveling faster
than vehicle 2, it will cover more distance during this time. We can say with certainty that the distance vehicle 1 travels plus the distance vehicle two travels will equal 480 miles. Written algebraically, 
. We
know that , so let's substitute 
for 
: 
. Since they both start at the same time, and meet
at the same time, the time for both vehicles is the same in our equation. Before proceeding further, factor the 
in the right side of the equation: 
(Equation B). Even
this equation, by itself, does not help us since there are infinite solutions. Let's call on Equation A to give us a hand. We will substitute equation A --solved for 
--into Equation B. When we do this, we have an equation that looks like this: 
. The time in this case is 4 hours; therefore, we substitute 4 for t giving 
. Now we have an equation we can solve because it only has on
e variable:
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Our original equation. |
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Combine like-terms. |
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Distribute the 4 to the right side. |
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Subtract 160 from both sides. |
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Divide both sides by 8. |
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This equation tells us that vehicle 2 was traveling at 40 mph. To calculate vehicle one's velocity, we substitute vehicles two's velocity into equation A, which was . Therefore,
. Vehicle 1 was traveling 80 mph.
Let's look at one more example before we apply this to accident reconstruction. How can we find the dimensions of a rectangle, knowing that its area is 50 square inches and its perimeter is 30 inches? Our first equation is that for area,
length times width (
). In this case, we know the area is 50 square inches, so 
. Our second equation deals with the perimeter; that is,
. Since we know that the perimeter is 30 inches,
. Let's simplify this last equation by dividing both sides, right now, by 2 (It is not necessary to perform this step, but doing so will make the math easier).
This gives us
. As with the first example, both equations, and 
, have an infinite number of solutions. But also similar to the
first example, we have two unknowns (l and w) and two equations. Solving for 
in the second equation gives us 
. One word here before we proceed: we can just as easily solve
for or 
, and in either equation. Now we substitute 
for 
, in the first equation:
. From here, we solve the equation algebraically. Distribute the 
in the right side of the equation to get 
At this point we must recognize
that our equation is second degree because w is squared. This tells us that we can use the quadratic equation, 
, to solve our equation. First, we must put our equation into standard form. In this case we get,
. Now we substitute our values into the quadratic equation and solve:
Note that the quadratic equation will always yield two answers. Which answer is
correct? Mathematically speaking, they are both correct. Conceptually however, only one answer is correct.
In real-world applications, we need to apply some reasoning to the solutions. If we accept that the width is 10, then the length, by the second equation, would be 
or 
=5. This would be a rectangle whose width is longer than its length. Our definitions of length and width specify that the length of an object is its longest
side; therefore, because of our definitions --and only because of our definitions --we reject this solution. When we try 5 for the width, then our length would be 15-5=10. This makes sense for our definitions of length and width. Is our solution correct?
Substituting 5 for 
, and 10 for 
into our original equations shows us that the area would be 50 square inches (10)(5), while the perimeter would be 30 inches
. Therefore, we have found the correct solution to our problem.
At this point, the reader should recognize that when we are presented with two unknown variables in an equation, we have an infinite number of solutions. However, if we have at our disposal two independent equations that describe the same two unknown variables, we are able, through substitution, to solve the system of equations. Other methods for solving a system of equations exist, but we will leave that for some other time.
Case Study #1
How can second-degree and systems of equations assist us with accident reconstruction? Imagine you are called to reconstruct a serious collision in which vehicle 1 strikes vehicle 2 in the rear. There is 100% overlap. The driver of vehicle 1 states that vehicle 2 was stopped on the highway. The driver of vehicle 2 states he was not stopped; in fact, he states he was traveling at the speed limit when vehicle 1 approached at a very high rate of speed and then struck his car. There are no witnesses to corroborate either of the driver's statements, and there is no physical evidence to suggest whether vehicle 2 was stopped or in motion. The only evidence available at the scene is distinct post-collision tiremarks from both vehicles, and roadway gouging that indicates the area of maximum engagement. Crush measurements were taken from both of these vehicles. From these crush measurements, an estimate of the energy dissipated damaging both vehicles is made. The following table depicts the only available information regarding this collision:
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Vehicle 1 |
Vehicle 2 |
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Mass |
90 Slugs* (2898 lbs) |
60 Slugs (1932 lbs) |
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Post Impact Velocity |
30 fps (20.4 mph) |
30 fps (20.4 mph) |
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Energy dissipated damaging both vehicles |
45,000 ft-lbs |
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(For a short discussion on mass, weight and slugs see Appendix A)
We will first look at this from the momentum aspect (we will assume there is no restitution, and both vehicles stayed together after impact). The equation for collinear momentum is 
. Because
the post impact velocities of both vehicles are the same, we rewrite the equation as 
. We substitute the information we know into the equation to get 
. Complete the
operations indicated in the right side of the equation to give us equation 1:
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Equation 1 |
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This should look familiar by now. We have an equation with two unknown variables; we need another independent equation to solve the system of equations. Our obvious choice is the conservation of energy. For this example, we will use the following equation:
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Energy Before The Collision |
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Energy Dissipated |
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Energy of V1 + Energy of V2 |
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Energy of V1 + Energy of V2 + Energy spent damaging both vehicles |
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= |
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Let's do some factoring and substitution before we proceed:
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Our original equation |
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Multiply both sides by 2 |
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Factor out the post velocities, which are the same for this collision. |
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Substitute known values |
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Perform operations in the parentheses. |
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Complete remaining operations (Equation 2) |
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Now we have a second equation to complete a system of equations. Before we do this, we must rearrange one of the equations for substitution into the other, by explicitly solving for one variable. We will do this with equation one, because
as a first-degree equation it will be easier. We now arbitrarily choose which variable to solve for;
we will solve for
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Equation 1 |
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Subtract |
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Divide both sides by 60 (Equation 3) |
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We now substitute equation 3 into equation 2:
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Our new equation with only one variable |
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Square the binomial in the left side. (See Appendix B for a short discussion on squaring a binomial.) |
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Distribute 60 |
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Combine like terms |
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Arrange the equation in standard form |
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from both sides.
from the left side
is an irrational number, which can only be approximated).
from both sides.
