Jermaine Dixon

Mathematical Modeling - Spring 2001

Hello! My project deals with
two scenarios that I encounter whenever I am behind the wheel of my car:
1)
The distance needed to safely enter onto a street without causing others
to slow down. and 2) The necessary
distance to accelerate the to flow of freeway traffic.

First, I wanted to generate a formula
that would represent the distances for both cars. Car A is the car
that approaches Car B, which wants to enter traffic from rest. I
began by using the formula for distance: D = r*t, where D is distance,
r is rate, and t is time. For Car A, I assumed that its rate was
a constant, 40 mph. Since the rate is constant, the formula for Car
A would be 40t. However, I had to consider the distance between the
cars once Car B wants to enter traffic. I use *s* to represent
this distance. Thus, my formula for the distance of Car A is 40t
- s.

Finding the distance formula for Car B was a little more tricky. Since its starts from rest and has to accelerate, I assumed that its rate would not be constant. Additionally, I sketched a graph of what I thought the rate would look like:

This graph did not give me formula, right away. Fortunately, from Calculus I remembered the relationship between graphs and their derivatives. This graph's derivative is very similar to this:

A parabola! Someone has already figured out this formula! [ y = a(x - h)^2 +k ] After some algebra, I found this equation to be:

Scary! At this point I should
state what *T* represents in the above equation. Where *t
*is
for time in general, *T* is the amount of time necessary for Car B
to acceleraterate to 40 mph.

Now, the equation that I found depicts
the derivative of my rate. To get the formula that I needed, I had
find the *antiderivative*, or integrate, the above equation.
Which gave me...

-320/3*t^3/(T^3)+160*t^2/(T^2)

...as a formula for the rate of Car B. The distance formula for Car B is

Next, I graphed the plots for both distances and evaluated them.

This is the distance graph for Car
A (blue) and Car B (red). The value of *s*, the distance between
the cars when Car B enters, is 30 ft. *T *is 5 sec. As
anyone can see, these paths cross. This translates into a crash.

I substituted values for *s*
until I found the smallest distance where the paths did not cross.

This is the distance graph where
*s
*=
47 ft, with
*T *still at 5 sec. From this information I conclude
that 47 ft is the minimal distance required between the car driving (Car
A) and the car entering traffic from rest (Car B) so that there will not
be an accident.

The 2nd scenario, merging onto the
freeway involves two cars (again), Car A and Car B. Car A is traviling
at a constant rate, which is assumed to 60 mph. Car B enters the
freeway from the onramp, in motion, and still accelerating. Another
assumption that I made is that the paths both Car A and Car B will intersect
at some point. Since both cars begin at a distance away from their
intersection, I assigned to variables to represent that distance. *p
*is
Car A's distance from the intersection and *q *is Car B's distance
from their intersection.

Aside from these two new variables, the process of obtaining the distance formulas for the cars was similar to the idea in the first Scenario. Car A's constant travel rate of 60 mph was simply substituted into the distance formula (D = r*t) giving me 60t - p, where p is the distance from the intersection of both cars.

For Car B, I found its rate by performing
a *phase shift* to my previous acceleration graph.

My reasoning for the phase shift
relies on the assumption that Car B is already in motion and accelerating,
which impies that at *t *= 0 the acceleration does
not equal zero. Again, using the formula
for parabolas (y = a(x - h) + k) I generated a formula for the above graph,

-480*1/(T^2*(-4+T))*(x-1/2*T+2)^2+120/(T-4)

which appears more complicated that
it is. Remember that *T* represents the time needed for Car
B to accelerate to 60 mph. To find Car B's rate, I again had to integrate
the acceleration formula.

-160*t^3/(T^2*(-4+T))-240*(4-T)*t^2/(T^2*(-4+T))-480*(-1/2*T+2)^2*t/(T^2*(-4+T))+120*t/(-4+T)

Multiply times *t *and subtract
*q*
(Car B's distance from the intersection of both cars),

t*(-160*t^3/(T^2*(-4+T))-240*(4-T)*t^2/(T^2*(-4+T))-480*(-1/2*T+2)^2*t/(T^2*(-4+T))+120*t/(-4+T))-q

and we have a distance formula for Car B.

The graph below is of the distance formulas of Car A and B,

where *p *(Car A's dist. from
their intersection) = 71 ft, *q *(Car B's dist. from their intersection)
= 131 ft, and *T *(the time required for Car B to accelerate to 60
mph) = 9 sec. These were the minimum values that I found. Thus,
I conclude that if Car B has at 131 ft it can accelerate up to 60 mph,
in 9 sec, and enter the freeway safely so long as there is not a
Car A that is less that 71 ft from their intersection at *T *= 0.

That's it!

For questions or comments, please
e-mail me at

jldixon@clunet.edu