Civil Engineering - Highway Engineering - Discussion
Discussion Forum : 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
Discussion:
10 comments Page 1 of 1.
Supriya Marik said:
1 year ago
C = 1000 V/S.
= 1000 * 24/(19+6)
= 960.
≈ 1000 (Answer C).
= 1000 * 24/(19+6)
= 960.
≈ 1000 (Answer C).
Parth said:
5 years ago
N = 1000v/L+C.
= 24000/25,
=960.
= 24000/25,
=960.
Parul verma said:
5 years ago
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.
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.
Burhaan said:
5 years ago
Agree, the Correct answer is 960.
Uddipan said:
7 years ago
Yes @D. Laxmi.
If it were 25 instead of 24. The value would be 1000.
If it were 25 instead of 24. The value would be 1000.
(1)
Garry said:
8 years ago
No, the correct answer is 960 veh/hr.
(1)
D.Laxmi Narayana said:
9 years ago
If the Speed of the vehicle is 25 KM/Hr.
Then the ANSWER is 1000.
Then the ANSWER is 1000.
(2)
Dnyanesh said:
9 years ago
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.
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.
(5)
Chitranjan said:
1 decade ago
C = 1000 V/S.
Where V = Velocity of moving vehicle.
S = Stopping distance + Average length of vehicle.
C = 1000*24/(19+6).
= 1000 (approx.).
Where V = Velocity of moving vehicle.
S = Stopping distance + Average length of vehicle.
C = 1000*24/(19+6).
= 1000 (approx.).
(2)
Deep said:
1 decade ago
C = 1000V/s.
= 1000x24/19.
= 1263.50.
= 1000x24/19.
= 1263.50.
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