On a road, vehicles are moving with average velocity of 80 kmph. The traffic volume on road is _____ vehicle/km.
3636

Open in App

Solution

The correct option is A 3636
$s = 0.2 V_{b} + 6 = 0.2 \times 80 + 6 = 22 m$

$q = K V = \frac{1000 \times V}{s} = \frac{1000 \times 80}{22}$

$= 3636.36 ≃ 3636 V e h i c l e / k m$

Suggest Corrections

Similar questions

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. A vehicle is moving on a rough road in a straight line with uniform velocity.

Q. If on a particular stretch of road, vehicles are moving at a speed of

40, 45, 48, 54

and

60 k m p h

, then the space mean speed is ____kmph.

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

Join BYJU'S Learning Program

On a road, vehicles are moving with average velocity of 80 kmph. The traffic volume on road is _____ vehicle/km.3636

The correct option is A 3636s=0.2Vb+6=0.2×80+6=22m q=KV=1000×Vs=1000×8022 =3636.36≃3636Vehicle/km

On a road, vehicles are moving with average velocity of 80 kmph. The traffic volume on road is _____ vehicle/km.
3636

The correct option is A 3636
$s = 0.2 V_{b} + 6 = 0.2 \times 80 + 6 = 22 m$

$q = K V = \frac{1000 \times V}{s} = \frac{1000 \times 80}{22}$

$= 3636.36 ≃ 3636 V e h i c l e / k m$