On a two lane road 260 vehicles are observed in 1 km stretch.If both lanes share equal vehicles then average space headway between vehicles is m.
7.69

Open in App

Solution

The correct option is A 7.69
Traffic density = 260 vehicles /km for 2 lane
= 130 vehicles /km / lane
$∴ Traffic density = \frac{1000}{S} = 130 v e h / k m$
$\Rightarrow S = 7.69 m$

Suggest Corrections

Similar questions

Q. A traffic stream in particular direction of a two-lane road moving with a constant speed of 50 km/h. with an average headway of 2.52 s.The longitudinal distance between two consecutive vehicles is

Average free flow speed and the jam density observed on a road stretch are 60 km/h and 120 vehicles/ km, respectively. For a linear speed density relationship, the maximum flow on the road stretch (in vehicles/h) is

Q. 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

Q. For a road the free flow speed is 60kmph and jam density is 75veh/km. During a particular time, there are 20 vehicles on a lane over 1 km of road. The average speed at that instant is

Q. On a road, vehicles are moving with average velocity of 80 kmph. The traffic volume on road is _____ vehicle/km.

Join BYJU'S Learning Program

On a two lane road 260 vehicles are observed in 1 km stretch.If both lanes share equal vehicles then average space headway between vehicles is m.7.69

The correct option is A 7.69Traffic density = 260 vehicles /km for 2 lane = 130 vehicles /km / lane ∴Traffic density=1000S=130 veh/km ⇒S=7.69 m

On a two lane road 260 vehicles are observed in 1 km stretch.If both lanes share equal vehicles then average space headway between vehicles is m.
7.69

The correct option is A 7.69
Traffic density = 260 vehicles /km for 2 lane
= 130 vehicles /km / lane
$∴ Traffic density = \frac{1000}{S} = 130 v e h / k m$
$\Rightarrow S = 7.69 m$