Traffic engineering miscellaneous
- The speed-density (u-k) relationship on a single lane road with unidirectional flow is u = 70 – 0.7 k, where u is in km/hr and k is in veh/km. The capacity of the road (in veh/hr) is _______
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Capacity = velocity × density
⇒ c = u × k
= (70 – 0.7 k) × k
= 70 k – 0.7 k2∴ dc = 70 - 1.4k = 0 dk
∴ k = 50
∴ c = 70 × 50 – 0.7 × (50)2 = 1750 veh / hr
Correct Option: C
Capacity = velocity × density
⇒ c = u × k
= (70 – 0.7 k) × k
= 70 k – 0.7 k2∴ dc = 70 - 1.4k = 0 dk
∴ k = 50
∴ c = 70 × 50 – 0.7 × (50)2 = 1750 veh / hr
- A traffic survey conducted on a road yields an average daily traffic count of 5000 vehicles. The axle load distribution on the same road is given in the following table.
The design period of the road is 15 years, the yearly traffic growth rate is 7.5% and the load safety factor (LSF) is 1.3. If the vehicle damage factor (VDF) is calculated from the above data, the design traffic (in million standard axle load, MSA) is __________.
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Vehicle damage factor (VDF)
VDF = V1 
W1 
4 + V2 W2 4 + V3 W3 4 + V4 W4 4 + V5 W5 4 Ws Ws Ws Ws Ws
V1 + V2 + V3 + V4 + V5
Ws = Standard axle load = 80 kW = 8.2 tonnes VDF
= 4.98Axle load = 365 × A[ (1 + r)n - 1 ] × VDF × F.S γ = 365 × 5000 × 
1 + 7.5 15 - 1 
× 4.98 × 1.3 = 308.59 100 7.5 100 Correct Option: A
Vehicle damage factor (VDF)
VDF = V1 W1 4 + V2 W2 4 + V3 W3 4 + V4 W4 4 + V5 W5 4 Ws Ws Ws Ws Ws
V1 + V2 + V3 + V4 + V5
Ws = Standard axle load = 80 kW = 8.2 tonnes VDF
= 4.98Axle load = 365 × A[ (1 + r)n - 1 ] × VDF × F.S γ = 365 × 5000 × 1 + 7.5 15 - 1 × 4.98 × 1.3 = 308.59 100 7.5 100
- It was observed that 150 vehicles crossed a particular location of a highway in a duration of 30 minutes. Assuming that vehicle arrival follows a negative exponential distribution, find out the number of time headways greater than 5 seconds in the above observation ?
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Probability of time head way greater than ‘tsec’ is given by p(h ≥ t) = e–m
If v = hourly flow rate, thenm = v t 3600 p(h ≥ t) = e–vt / 3600 = -150 × 2 × 5 = 0.659254 e 3600
No. of time headway greater than ‘t’ sec = (no of observed time headway) × p (h ≥ t)
= 150 × 0.65924 = 98.88
Correct Option: B
Probability of time head way greater than ‘tsec’ is given by p(h ≥ t) = e–m
If v = hourly flow rate, thenm = v t 3600 p(h ≥ t) = e–vt / 3600 = -150 × 2 × 5 = 0.659254 e 3600
No. of time headway greater than ‘t’ sec = (no of observed time headway) × p (h ≥ t)
= 150 × 0.65924 = 98.88
- On a section of a highway the speed-density relationship is linear and is
given by v = 80 - 2 k ; 3
where v is in km/h and k is the veh/kem. The capacity (in veh/h) of this section of the highway would be
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Given , V = 80 - 2 k 3
Capacity of section, q = V × k= 80 - 2 k × k = 80k - 2 k2 .......(1) 3 3
For maximum q,
Differentiate q. wrt kdq = 0 dk ⇒ 80 - 4 k = 0 3
⇒ k = 60
Substitute in (1)q = 80 × 60 - 2 × 602 = 2400 veh / hr 3 Correct Option: B
Given , V = 80 - 2 k 3
Capacity of section, q = V × k= 80 - 2 k × k = 80k - 2 k2 .......(1) 3 3
For maximum q,
Differentiate q. wrt kdq = 0 dk ⇒ 80 - 4 k = 0 3
⇒ k = 60
Substitute in (1)q = 80 × 60 - 2 × 602 = 2400 veh / hr 3
- A student riding a bicycle on a 5 km one-way street takes 40 minutes to reach home. The student stopped for 15 minutes during this ride. 60 vehicles overtook the student (assume the number of vehicles overtaken by the student is zero) during the ride and 45 vehicles while the student stopped. The speed of vehicle stream on that road (in km/hr) is
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Velocity of bicycle = 5 = 12 km / h (40 - 15)
Speed of each vehicle when student is riding and when student stopped is same
= 60 /(40 - 15) = 45 / 15 S - 12 S
∴ Speed of vehicle, S = 60 km/hrCorrect Option: D
Velocity of bicycle = 5 = 12 km / h (40 - 15)
Speed of each vehicle when student is riding and when student stopped is same= 60 /(40 - 15) = 45 / 15 S - 12 S
∴ Speed of vehicle, S = 60 km/hr
