James R. Davis
Posted - 05/29/2008 : 5:18 PM
| Here is an example that shows just how powerful the quadratic equation solver can be.
Let us say that two cars, one traveling east while the other traveling west, have a head-on collision where there is a 100% overlap (meaning, full face hits full face rather than, say, driver's side hits driver's side).
Car 1, traveling eastbound, weighs about 2898 pounds and car 2, traveling westbound, weighs about 1932 pounds.
After the collision the two cars remain attached to each other and both move in an easterly direction at 14 ft/sec.
Crush damage determines that 145,800 ft-lbs of energy was dissipated doing damage to both cars.
Can you determine the speed each car was moving at the time of impact?
We will use two different equations and integrate them in an attempt to create a standard form resulting equation in order to then use the quadratic equation to resolve each vehicle's speed at impact. (Speeds will be positive when traveling eastbound.)
The first equation will deal with momentums, while the second will deal with energy.
Conservation of momentum means that the starting momentum equals the ending momentum.
Momentum is mass times velocity. Therefore,
m1v1 + m2v2 = m1v1' + m2v2'
2898 lbs = 90 slugs (2898 / 32.2)
1932 lbs = 60 slugs (1932 / 32.2)
Because both cars were moving at the same speed after the collision, we know that velocity v1' = v2', so we will call that v' and simplify the first equation:
m1v1 + m2v2 = v'(m1 + m2)
We then substitute known values:
90v1 + 60v2 = 14(90 + 60)
90v1 + 60v2 = 2100 Equation (1)
Energy is equal to 1/2 the mass times the velocity squared.
Conservation of energy means that the starting energy equals the ending energy plus energy lost in the collision.
Thus, we are able to create our second equation.
(m1/2)v1^2 + (m2/2)v2^2 = (m1/2)v1'^2 +(m2/2)v2'^2 + 145,800
Multiply both sides by 2
m1v1^2 + m2v2^2 = m1v1'^2 + m2v2'^2 + 291,600
Again, because both cars had the same speed after the collision, v1' and v2' are the same and we will call that v' and re-write:
m1v1^2 + m2v2^2 = v'^2(m1 + m2) + 291,600
Substituting known values:
90v1^2 + 60v2^2 = 14(90 +60)^2 + 291,600
90v1^2 + 60v2^2 = 14(150)^2 + 291,600
90v1^2 + 60v2^2 = 321,000 Equation (2)
Now let's solve for either v1 or v2 in the first equation. We use the first equation only because it is a first-degree, and therefore, easier to solve equation. We will solve for v2, but either unknown works just as well.
90v1 + 60v2 = 2100
60v2 = 2100 - 90v1
v2 = (35 - 1.5v1) Equation (3)
Now we integrate the two independent equations by substituting the value of v2 we just calculated into equation (2):
90v1^2 + 60v2^2 = 321,000
90v1^2 + 60(35 - 1.5v1)^2 = 321,000
We need to square the binomial.
(35 - 1.5v1)(35 - 1.5v1)
(1,225 -52.5v1 -52.5v1 +2.25v1^2)
(1,225 -105v1 +2.25v1^2)
Make it standard form:
(2.25v1^2 - 105v1 +1,225)
Plug it into our integrated equation:
90v1^2 + 60(2.25v1^2 -105v1 +1,225)^2 = 321,000
We distribute 60 on the left side:
90v1^2 + 135v1^2 - 6,300v1 + 73,500 = 321,000
225v1^2 - 6,300v1 + 73,500 = 321,000
Subtract 321,000 from both sides:
225v1^2 - 6,300v1 - 247,500 = 0
And now that we have a standard form equation we can use the quadratic to solve it (for v1) because we know that a=225, b=-6,300 and c=-247,500.
(See the graphic above.)
The two values for v1, both mathematically correct, are 50 fps and -22 fps.
If we now substitute those values into equation (3) we learn that the values for v2 are -40 fps and 68 fps respectively.
In other words, we have determined that EITHER car 1 was traveling eastbound at 50 fps while car 2 was traveling westbound at 22 fps at the time of impact, or that car 1 was traveling westbound at 40 fps while car 2 was traveling eastbound at 68 fps.
Since the problem stated that car 1 was traveling eastbound, yet the second solution shows that it was traveling eastbound at 40 fps, the second viable solution is shown to be invalid.
We now know that car 1 was traveling eastbound at 50 fps when it collided with car 2 which was traveling westbound at 22 fps.