A two-lane urban road with one-way traffic has a maximum capacity of 1800 vehicles/hour. Under the jam condition, the average length occupied by the vehicles is 5 m. The speed versus density relationship is linear, For a traffic volume of 1000 vehicles/hour, the density (in vehicles/km) is

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Solution

The correct option is D 67
Jam density per lane
$K_{j} = \frac{1000}{5} = 200 v e h i c l e / k m$
Maximum capacity per lane
$= \frac{1800}{2} = 900 v e h / h r$
$q_{m a x} = \frac{V_{f} K_{j}}{4}$
$V_{f} = \frac{4 \times 900}{200} = 18 k m / h r$
$q = V_{f} K - \frac{V_{f}}{K_{j}} K^{2}$
Traffic Vol. per lane = 500
$500 = 18 K - \frac{18}{200} K^{2}$
$K^{2} - 200 K + \frac{100000}{18} = 0$
$K = 33.33 v e h / k m$
Hence, density for two lane = $2 \times 33.33 =$ 66.67 veh/km

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A two-lane urban road with one-way traffic has a maximum capacity of 1800 vehicles/hour. Under the jam condition, the average length occupied by the vehicles is 5 m. The speed versus density relationship is linear, For a traffic volume of 1000 vehicles/hour, the density (in vehicles/km) is

The correct option is D 67Jam density per lane Kj=10005=200vehicle/km Maximum capacity per lane =18002=900veh/hr qmax=VfKj4 Vf=4×900200=18km/hr q=VfK−VfKjK2 Traffic Vol. per lane = 500 500=18K−18200K2 K2−200K+10000018=0 K=33.33veh/km Hence, density for two lane = 2×33.33=66.67 veh/km