Civil Engineering - Highway Engineering - Discussion

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

Answer: Option C

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.

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