# Civil Engineering - Highway Engineering - Discussion

### Discussion :: Highway Engineering - Section 3 (Q.No.31)

31.

If the velocity of moving vehicles on a road is 24 km/per hour, stopping distance is 19 metres and average length of vehicles is 6 metres, the basic capacity of lane, is

 [A]. 500 vehicles per hour [B]. 700 vehicles per hour [C]. 1000 vehicles per hour [D]. 1250 vehicles per hour

Explanation:

No answer description available for this question.

 Deep said: (Aug 21, 2014) C = 1000V/s. = 1000x24/19. = 1263.50.

 Chitranjan said: (Nov 13, 2014) C = 1000 V/S. Where V = Velocity of moving vehicle. S = Stopping distance + Average length of vehicle. C = 1000*24/(19+6). = 1000 (approx.).

 Dnyanesh said: (Nov 20, 2016) There is a small mistake @Chitranjan C = 1000 V/S. Where V = Velocity of moving the vehicle. S = Stopping distance + Average length of the vehicle. C = 1000 * 24/(19 + 6). = 960.

 D.Laxmi Narayana said: (Dec 9, 2016) If the Speed of the vehicle is 25 KM/Hr. Then the ANSWER is 1000.

 Garry said: (Oct 15, 2017) No, the correct answer is 960 veh/hr.

 Uddipan said: (Nov 25, 2018) Yes @D. Laxmi. If it were 25 instead of 24. The value would be 1000.

 Burhaan said: (Jun 24, 2020) Agree, the Correct answer is 960.

 Parul Verma said: (Aug 11, 2020) Capacity flow or maximum flow = (V*K)÷ 4 V- velocity. K- jam density. K= 1000÷ s. S- spacing between the vehicle. K = 1000÷6 =166.67. Maximum flow= (24*66.67)÷ 4 = 1000.

 Parth said: (Nov 20, 2020) N = 1000v/L+C. = 24000/25, =960.