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ki77a [65]
3 years ago
6

There are 10 vehicles in a queue when an attendant opens a toll booth. Vehicles arrive at the booth at a rate of 4 per minute. T

he attendant opens the booth and improves the service rate over time following the function u(t) = 1.1 + .30t, where u(t) is in vehicles per minute and t is in minutes. I need to find the maximum queue length. I included my solution below for the total delay.
u(t) = 4 when t = (4-1.1)/0.3 = 9.67mins
(9.67 to t) ? (1.1+0.3t)dt = 10
? 1.1t + 0.15t2 ](9.67 to t)= 10
? t = 11.97mins
So, delay = 11.97 – 9.67 = 2.30mins
Engineering
1 answer:
dmitriy555 [2]3 years ago
3 0

Answer:

as slated in your solution, if delay time is 2.30 mins, hence 9 vehicle will be on queue as the improved service commenced.

Explanation:

4 vehicle per min, in 2 mins of the delay time 8 vehicles while in 0.3 min average of 1 vehicle join the queue. making 9 vehicle maximum

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- Secondary wastewater treatment: treatment processes remove waste organic ( which are once living/biological) material from wastewater, usually by the use of a biological treatment process.

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Secondary wastewater treatment processes remove waste organic ( which are once living/biological) material from wastewater, usually by the use of a biological treatment process. The goal of secondary treatment is the further treatment of the effluent from primary treatment in order to get rid of the residual organics and suspended solids.

Advanced wastewater is quite in between. Tertiary treatment isn't needed at all wastewater treatment plants, and the place it is needed, it may be different from one plant to another, depending upon the type of water contamination that must be removed. Advanced treatment processes can be combined with primary or secondary treatment or used instead of secondary treatment.

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pV¹°¹ = constant

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We know,

V1¹°¹ = p2V2¹°¹

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8 0
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