Optimizing Traffic Flow with Collision-Avoidance Automated Control Systems

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Current freeways can sustain approximately 2400 vehicles per lane per hour at a speed of 96 km/h before traffic flow becomes turbulent. The average spacing between vehicles at this density is calculated to be about 40 meters, including both vehicle length and spacing. Collision-avoidance automated control systems can reduce the required spacing to 9.2 meters, allowing for a significant increase in traffic capacity. With this reduced spacing, a lane can accommodate around 6957 vehicles per hour at the same speed. Optimizing traffic flow with these systems could greatly enhance roadway efficiency.
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According to a Scientific American article, current freeways can sustain about 2400 vehicles per lane per hour in smooth traffic flow at 96 km/h (60 mi/h). With more vehicles the traffic flow becomes "turbulent" (stop-and-go).

a) If a vehicle is 4.6 m (15ft) long on the average, what is the average spacing between vehicles at the above traffic density?

b) Collision-avoidance automated control systems, which operate by bouncing radar or sonar signals off surrounding vehicles and then accelerate or brake the car when necessary, could greatly reduce the required spacing between vehicles. If the average spacing is 9.2m (two car lengths), how many vehicles per hour can a lane of traffic carry at 96 km/h?

I don't know how to approach this problem. Please help. Thanks!
 
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Firstly, look at the data you have been given. Traffic flows at 96 km/h. It means that, in 1 hour, 96km stretch of traffic is cleared. Also, 2400 vehicles per lane can be sustained in 1 hour. It means that 2400 vehicles must occupy a minimum of 96km. If the no. of cars are increased, then at the same speed, more no. of vehicles occupy lesser distance causing congestion.

Here, the key to the problem is that u need to think in terms of length only.

Now, 96km = 96,000m. If this is what 2400 vehicles occupy [lengthwise only], 1 car will occupy \frac{96000}{2400}~m. Which equals to 40m. Each car, therefore can occupy 40m. But, the length of the car is 4.2m. Which means, it can also have a free space of 40 - 4.6 = 35.4m

For the 2nd problem, try to think of it yourself. Here, the average spacing is 9.2m as opposed to 35.4m in the 1st part.
 
Thanks! That was a great explanation!
I got 6957vehicles/h for part b.
 
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