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
500 vehicles per hour
700 vehicles per hour
1000 vehicles per hour
1250 vehicles per hour
Answer: Option
Explanation:
No answer description is available. Let's discuss.
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).

Parth said:   5 years ago
N = 1000v/L+C.
= 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.

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.
(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.
(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.
(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.).
(2)

Deep said:   1 decade ago
C = 1000V/s.

= 1000x24/19.

= 1263.50.

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