A fourlane highway has a normal capacity of 1800 vehhrlane I

A four-lane highway has a normal capacity of 1800 veh/hr/lane. In the southbound direction, a vehicle disablement on the roadway shoulder occurs at 4:30 p.m. Due to rubbernecking, the capacity in the southbound direction is reduced to 1200 veh/hr/ lane at this time. At 4:45 p.m., the disabled vehicle is removed from the shoulder and the capacity increases to 1500 veh/hr/lane. At 5:00 p.m. the roadway capacity returns to its full value of 1800 veh/hr/lane. From 4:30 p.m. until the queue clears the traffic flow is 1600 veh/hr/lane. Determine the following: Average delay per vehicle Maximum queue length Average queue length in the southbound direction

Solution

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