Answer:
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.