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Traffic flow on a one-lane road section (no passing is allowed) has the capacity of 1,000 vehicles/hour and free-flow speed o

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Answer #1

Answer;

Given;

free flow speed(v)= 40km/hr

capacity(q)= 1000veh/hr

we know from the fundamental equation of traffic flow,

q=k×v

1000=k×40

k =1000/40   =25veh/km

Now,

(a)Jam Density by Greenshield’s model

q =k×v / 4

q=25×40/ 4    =250veh/hr

So, Jam density = 250 veh/hr

Now, Maximum density= Half of Jam density, i.e

Km (Maximum Density or critical Density) =kj(Jam Density) /2

So,Critical Density k=250/2 =125 veh/km

q max=kj vf /4

        =250×40/4

       qmax=2500 veh/hr

qm (2500) E- Flow(9) - - - - ok isterik) kj (250)

(b)Length of platoon when the gate was opened

Lets first characterize the approach and platoon conditions

APPROACH CONDITIONS PLATOON CONDITIONS

qa= 640 veh/hr qp= 0 veh/hr

u =40 km/hr u=0

k= 20 veh/km     k=kj =250 veh/km

u=(0-640) / (250-20) = -2.782 km/hr

Length of the que after 3 minutes = u * t

=2.782 * (3/60)

Length of que   =0.1391 km

(c) Time for the platoon to dissipate after opening of the gate

Again lets characterize the platoon and release conditions

PLATOON CONDITIONS RELEASE CONDITIONS

q=0 veh/hr q=qm=2500 veh/hr

v=0 km/hr v=40 km/hr

k=250 veh/km k=km(no congestion) = 250/2 =125 veh/km

u=(0-2500) / (250-125) =-20 km/hr

U=-20-(-2.782)= 17.218 km/hr

t=L/U

t=0.1391/17.218 = 0.008 hr = 0.417 min

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