Question

Consider a freeway with two lanes and capacity of 2000 veh/h/lane, jam density of 60 veh/km/lane and a triangular fundamental diagram. Traffic is moving with the flow of 3000 veh/h and the space mean speed of 100 km/h. At 10:00 a.m., a traffic crash occurs that blocks one lane for half an hour. Using shockwave analysis, answer the following questions. a. Draw the space-time diagram with shockwaves and traffic states (trajectories not needed). [2 marks] b. At time t = 10:30 a.m., Jack enters the freeway via an upstream on-ramp, which is 20 km away from the accident location. Does Jack experience congestion? If yes, what time does Jack start driving at a lower speed? [3 marks]

Answer 1

We can put the data given to us in terms of a table with arrival rate, # of arrivals in 15 minutes, departure rate and number of departures. We can calculate Cumulative arrivals and departures and then calculate queue length. So we have the following

From 10 to 10:30 the vehicles are arriving at 3000 veh/hour (i.e over two lanes). From 10 to 10:30, one lane is blocked, so the availble lane will operate at maximum dischare capacity i.e 2000 veh/hour.

After 10:30, both lanes are open, and since there is a queue built up, the discharge capacity can be assumed to be maximum over two lanes i.e 4000 veh/hour over two lanes.

So we have

So let us plot this on a graph showing Cumulative Arrivals and Departures

Maximum queue occurs at time period starting at 10:15am. So it actually happens between 10:15am and 10:30am.

This maximum queue is 500vehicles. It then drops to 125 vehicles after 2nd lane is opened and then dissipates completely

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So at 10:00 am a queue starts forming and this is represented by a backward forming shockwave which indicates that the tail of the queue is coming back.

The speed of this shockwave can be calculated as follows

When the queue starts forming, we have a backward forming shockwave.

arriving flow upstream = 3000 veh/hr with a speed of 100 km/hour

so density upstream = 3000/100 = 30 veh/km but this is over 2 lanes, so over one lane it is 15 veh/km/lln.

So at free flow it is a speed of 100 km/hr with a density of 15 veh/km/ln.

When queue starts forming, it gets to jam density, and at the bottle neck, the density increases and this new density can be calculated as

Speed of the backward shockwave (i.e at which queue is forming and travelling backward) =

where

q_approach = 3000 veh/hr this is over two lanes, so its basically 1500 veh/hr

k_approach = 15 veh/mi/lane

k_jam = 60 veh/mi/lane

So speed of backward shockwave = u_wr = (1500-0)/(15-60) = -1500/145 = -33.333 km/hour

so after 30 minutes the length of the queue = Speed of queue shockwave * time

= =33 km/hour * 0.5 hours = 16.66 km

jam density = 60 veh/km/ln

So number of vehicles queued up = 60 * 16.66 =

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