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This one is a really tough one for me too?

According to a Scientific American article (May, 1990), current freeways can sustain about 2350 vehicles per lane per hour in smooth traffic flow at speed 97 . Above that figure the traffic flow becomes "turbulent" (stop-and-go).

there are two parts to the question.
PART 1
If a vehicle is of length 4.7 on the average, what is the average spacing between vehicles at the above traffic density?

PART 2
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 a distance 9.1 , how many vehicles per hour can a lane of traffic carry at speed 97 ?

1) If 2350 vehicles pass any given point in one hour, that means you’ve got

(2350 vehicles/hour) / (60 minutes/hour) / (60 seconds/minute) = 0.65 vehicles/second

At 97 m/s, that means each vehicle needs a total of

97m/s * 0.65s = 63.3m.

Since the average vehicle is 4.7m, subtract that from the total space, to get 58.6m of space between vehicles.

2) Same process as the previous problem, just in reverse. 9.1m of space plus 4.7m for the vehicle is 13.8m. At 97 m/s

(13.8m) / (97 m/s) = 0.142s

A vehicle will pass a given point every 0.142 seconds. There are 3600 seconds in an hour so:

(3600 s/hr) / (1 vehicle/0.142s) = 25352 vehicles/hr

This entry was posted on Monday, September 28th, 2009 at 2:31 am and is filed under automated traffic system. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

One Response to “This one is a really tough one for me too?”

  1. 1) If 2350 vehicles pass any given point in one hour, that means you’ve got

    (2350 vehicles/hour) / (60 minutes/hour) / (60 seconds/minute) = 0.65 vehicles/second

    At 97 m/s, that means each vehicle needs a total of

    97m/s * 0.65s = 63.3m.

    Since the average vehicle is 4.7m, subtract that from the total space, to get 58.6m of space between vehicles.

    2) Same process as the previous problem, just in reverse. 9.1m of space plus 4.7m for the vehicle is 13.8m. At 97 m/s

    (13.8m) / (97 m/s) = 0.142s

    A vehicle will pass a given point every 0.142 seconds. There are 3600 seconds in an hour so:

    (3600 s/hr) / (1 vehicle/0.142s) = 25352 vehicles/hr
    References :

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