Answer to Question #92157 in Algebra for lee

Question

Assumptions:
1. All cars are approximately the same length (average = 4.2m).
In dense, uniform traffic flow, all cars maintain the same “safe distance” apart.
This “safe distance” is the stopping distance required for a car travelling at V km/h. This distance is composed of two parts –
Reaction Distance, the distance the car travels as the driver reacts to the situation and applies the brake, and Braking Distance, the distance the car travels once the brake has been applied.
The “safe distance” can be reliably obtained from the table given on the next page based on the Ministry of Transportation Code used in England.
Number of cars/second=(m/second)/(m/car)
Show how the formula for traffic flow was developed. Include in your response any assumptions and their associated effects.
Velocity (Km/h) RD m BD m
10 1.905 0.600
20 3.810 2.400
30 5.715 5.400
40 7.620 9.600
50 9.525 15.000
60 11.430 21.600
70 13.335 29.400
80 15.240 38.400
90 17.145 48.600
100 19.050 60.000

Answer 1

Plot a graph with velocity vs. reaction distance:

We see that this is a linear function, and human reaction time is the same for any point on the graph:

$$t=\frac{RD}{v}=\frac{15.240}{80\cdot1000/3600}=0.686\text{ s}.$$

Therefore, the reaction distance (m) for any velocity (m/s) can be found as:

$$RD=0.686v\text{ m}.$$

The braking distance is a distance a car traveled while it decelerated. Thus, for any point "standard" maximum safe deceleration can be calculated as

$$a=\frac{v^2}{2BD}=\frac{40^2}{2\cdot9.6}=6.430\text{ m/s}^2.$$

Therefore, the braking distance is

$$BD=\frac{v^2}{12.86}\text{ m}.$$

The traffic flow is the number of cars per second. Between the rear bumper of any previous car and the front bumper of any following car there is the safe distance:

$$SD=RD+BD=0.686v+\frac{v^2}{12.86}\text{ m}.$$

Hence, per 1 second the traffic flow is the velocity over sum of the safe distance and the length of a car:

$$TF=\frac{v}{SD+L}=\frac{v}{0.686v+\frac{v^2}{12.86}+4.2}.$$