Question

. A four-lane arterial (with two lanes in each direction) has a capacity of 1000 vehicleshourlane. Due to some road repair work, the maximum capacity of two lanes in the northbound direction is reduced to 1100 veh/hr from 2000 veh/hr for a short length of 200 ft. If the total traffic upstream of the repair site in the northbound lanes is reasonably at 1500 vehhr, find (a the mean speed of traffic through the repair site and (b) the stead rate at which the queue approaching the repair site grows. Assume the jam density is 200 vehicles/milenane.

Answer 1

Note : it is OR problem not for Advanced physics, in future do not post such problem to AP, it may not be answered by anyone.

from 4:40 pm to 4:45 pm flow rate = 1600 veh/hr

clearing rate = 1200 veh/hr

accumulation = (1600-1200)/4 = 100 veh

on an average there are 100 vehicles in queue between 4:30 to 4:45

delay per vehicle = 60*100/1200 = 5 mts.

from 4:45 to 5:00 pm

inward flow rate = 1600 veh/hr

output = 1500 veh/hr

accumulation = (1600-1500)/4 = 25 veh

delay per vehicle = 60*25/1500 = 1mt

100 vehicle have delay of 5mts

25 vehicles have delay of 1 mt

average delay per vehicle = (100*5+25*1)/125

                                          = 4.2 mts

maximum queue length = 100 veh

average queue length = (100*15+25*15)/30 =62.5 veh