5 500 18345 vehicles assume a headtohead distance for

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Unformatted text preview: The time spent in the queue for the 500th vehicle is the maximum delay time  tmax = 35 – 7.5 = 27.5 minutes Maximum queue length occurs at the 35th minute Total vehicles that have arrived up to this time are: 35 * 66.7 = 2334.5 veh Tot vehicles serviced (left the queue at this point) are: 25*(35 ­15) = 500 vehicles The difference is the max queue length in vehicles = 2334.5 – 500 = 1834.5 vehicles Assume a head ­to ­head distance for vehicles in queue (mine is 20 ft) The total queue length in miles is 1834.5 * 20 = 36690ft/5280 (ft/mi) = 6.95 miles To find average delay, you need to find total delay first (the area...
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This document was uploaded on 02/25/2014 for the course CEE 207 at Lehigh University .

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