A toll booth on a turnpike is open from 8:00 AM until midnight. Vehicles start arriving at 7:45 AM at a uniform rate of 6 veh/min. At 8:15 AM the rate drops to 2 veh/min. If vehicles are processed at a uniform rate of 6 veh/min, determine when the queue will end, the total delay, the maximum queue length, and the longest vehicle delay.

The vehicles start arriving at 7:45 AM at a rate of 6 veh/min.

then, until the 8:00 AM that the vehicles start being processed, the amount of vehicles accumulated is 6veh/min*15min = 6*15 vehicles = 90 vehicles.

At 8:00 am, when the vehicles start to being processed, we have a queue of 90 vehicles, and at this point, each minute 6 vehicles arrive and also 6 vehicles are processed, so the queue does not change.

At 8:15 AM, the rate at which the vehicles arrive is equal to 2 veh/min.

then, from this hour we have that the number of vehicles in the queue can be described by the equation:

V (t) = 90 + 2*t - 6*t

where t is the number of minutes that passed since the 8:15 AM

this means: we have an initial number of 90 vehicles, and every minute 2 new arrive and 6 are processed.

The queue will end when we have that V(t) = 0

this is:

90 - 4t = 0

t = 90/4 = 22.5 mins.

This means that the queue will end at:

8:15 am + 22.5mins = 8:37.5 Am

You also ask for the longest delay, I guess this is how much a vehicle must wait until it is processed.

knowing that 6 vehicles are processed per minute, the car in the last place (car number 90) must wait 15 minutes, because of 6*15 = 90

so the longest delay is 15 mins.

and this is the same for the first car that arrives at 7:45 AM, this car also needs to wait for 15 minutes.

e/c, where e is its total energy