Question 1. The critical flow ratios for a three-phase signal are found to be 0.30, 0.25, and 0.25. The total time lost in the cycle is 10 s. Pedestrian crossings at the junction are not significant. The respective Green times (expressed in seconds and rounded off to the nearest integer) for the three phases are [2016 : 2 Marks, Set-II]
(a) 34, 28 and 28
(b) 40, 25, and 25
(c) 40, 30 and 30
(d) 50, 25, and 25
Ans: (a)
Explanation:
Use Webster's method to find the optimum cycle and allocate effective green times proportionally to critical flow ratios.
Sum of critical flow ratios, Y = 0.30 + 0.25 + 0.25 = 0.80.
Lost time, L = 10 s.
Webster's optimum cycle, C0 = (1.5L + 5) / (1 - Y) = (1.5×10 + 5) / (1 - 0.80) = (15 + 5) / 0.20 = 100 s.
Effective green available = C0 - L = 100 - 10 = 90 s.
Allocate effective green proportionally: gi = (yi / Y) × (C0 - L).
g1 = (0.30 / 0.80) × 90 = 33.75 ≈ 34 s.
g2 = (0.25 / 0.80) × 90 = 28.125 ≈ 28 s.
g3 = (0.25 / 0.80) × 90 = 28.125 ≈ 28 s.
Rounded green times: 34 s, 28 s, 28 s (option a).
Question 2. If the total number of commercial vehicles per day ranges from 3000 to 6000, the minimum percentage of commercial traffic to be surveyed for axle load is [2016 : 1 Mark, Set-II]
(a) 15
(b) 20
(c) 25
(d) 30
Ans: (a)
Explanation:
As per standard practise (IRC recommendations), when daily commercial vehicles are between 3,000 and 6,000, a minimum sample of 15% of commercial vehicles should be surveyed for axle load studies. This ensures a representative axle-load sample while keeping the survey effort reasonable.
Question 3. While traveling along and against the traffic stream, a moving observer measured the relative flows as 50 vehicles/hr and 200 vehicles/hr, respectively. The average speeds of the moving observer while traveling along and against the stream are 20 km/hr and 30 km/hr, respectively. The density of the traffic stream (expressed in vehicles/km) is________. [2016 : 2 Marks, Set-I]
Solution:
Method-I
Let q = traffic flow (veh/hr), k = traffic density (veh/km).
When observer travels along the stream at V₁ = 20 km/h, relative flow observed = q - k V₁ = 50 veh/hr. ...(1)
When observer travels against the stream at V₂ = 30 km/h, relative flow observed = q + k V₂ = 200 veh/hr. ...(2)
Subtract (1) from (2): (q + 30k) - (q - 20k) = 200 - 50 → 50k = 150 → k = 3 veh/km.
Thus density k = 3 vehicles/km.
Method-II
Consider an L km stretch and count vehicles overtaking and being overtaken while moving along and against. The same pair of linear equations (q - kV₁ and q + kV₂) result, leading to k = 3 veh/km.
Question 4. Inaone-lane one-way homogeneous traffic stream, the observed average headway is 3.0s. The flow (expression vehicles/hr) in this traffic stream is __________ [2016 : 1 Mark, Set-I]
Solution: Given: Average headway, h = 3.0 s.
Flow q (veh/hr) = 3600 / h = 3600 / 3.0 = 1200 veh/hr.
Quesion 5. The relation between speed u(in km/h) and density k (number of vehicles/km) for a traffic stream on a road is u = 70 - 0.7k. The capacity on this road is ________ vph (vehicles/hour). [2015 : 2 Marks, Set-II]
Solution: Traffic volume q = u k = (70 - 0.7k) k = 70k - 0.7 k2.
To find capacity (maximum q), set derivative dq/dk = 0:
dq/dk = 70 - 1.4k = 0 → k = 70 / 1.4 = 50 veh/km.
Capacity q_max = 70×50 - 0.7×50² = 3500 - 0.7×2500 = 3500 - 1750 = 1750 veh/hr.
Thus the capacity is 1750 vph.
Question 6. The following statements are made related to the lengths of turning lanes at signalised intersections
(i) 1.5 times the average number of vehicles (by vehicle type) that would store in turning lane per cycle during the peak hour.
(ii) 2 times the average number of vehicles (by vehicle type) that would store in turning lane per cycle during the peak hour.
(iii) Average number of vehicles (by vehicle type) that would store in the adjacent through lane per cycle during the peak hour.
(iv) Average number number of vehicles (by vehicle type) that would store in all lanes per cycle during the peak hour.
As per the IRC recommendations, the correct choice for design length of storage lanes is [2015 :1 Mark, Set-Il]
(a) Maximum of (ii) and (iii)
(b) Maximum of (i) and (iii)
(c) Average of (i) and (iii)
(d) Only (iv)
Answer: (c)
Solution: IRC recommends that the design storage length for turning lanes be taken as the average of (i) and (iii).
Explanation:
(i) provides a factor to allow some margin over the expected stored vehicles in the turning lane; (iii) accounts for the adjacent through lane demand. Averaging these two gives a balanced design length to accommodate turning queues without excessive extra pavement.
Question 7. Which of the following statements CANNOT be used to describe free flow speed (uf) of a traffic stream? [2015 : 1 Mark, Set-I]
(a) uf is the speed when flow is negligible
(b) uf is the speed when density is negligible
(c) uf is affected by geometry and surface conditions of the road
(d) uf is the speed at which flow is maximum and density is optimum
Ans: (d)
Explanation:
Free flow speed is the speed of vehicles when interactions between vehicles are negligible (very low flow and low density) and is influenced by road geometry and surface conditions. It is not the speed at which flow is maximum; maximum flow occurs at some lower speed when density and interactions are non-negligible. Hence statement (d) is incorrect.
Question 8. A pre-timed four phase signal has critical lane flow rate for the first three phases as 200, 187 and 210 veh/hr with saturation flow rate of 1800 veh/hr/lane for all phases. The lost time is given as 4 seconds for each phase. If the cycle length is 60 seconds, the effective green time (in seconds) of the fourth phase i s ________ . [2014 : 2 Marks, Set-II]
Solution: Flow rates for the first three phase are given as
q1 = 200 veh/hr
q2 = 187 veh/hr
and q3 = 210 veh/hr
Saturation flow rate s = 1800 veh/hr/lane for each phase.
Lost time per phase = 4 s → Total lost time L = 4 × 4 = 16 s.
Cycle length C = 60 s → Effective green available = C - L = 60 - 16 = 44 s.
For each phase, gi = (qi × C) / s. The sum of gi over all phases must equal C - L.
Thus (C / s) × (q1 + q2 + q3 + q4) = C - L → q1 + q2 + q3 + q4 = s × (1 - L/C).
Compute s × (1 - L/C) = 1800 × (1 - 16/60) = 1800 × 0.733333... = 1320 veh/hr.
Sum of known flows = 200 + 187 + 210 = 597 veh/hr.
Therefore q4 = 1320 - 597 = 723 veh/hr.
Effective green of phase 4, g4 = (q4 × C) / s = (723 × 60) / 1800 = 24.1 s (approximately).
Hence effective green time of the fourth phase ≈ 24.1 s.
Question 9. On a section of a highway the speed- density relationship is linear and is given by v = [80 - (2/3) k]; where v is in km/h and k is in veh/km. The capacity (in veh/h) of this section of the highway would be [2014 : 2 Marks, Set-II]
(a) 1200
(b) 2400
(c) 4800
(d) 9600
Answer: (b)
Solution:
Method-I
Traffic flow q = v k = k[80 - (2/3)k] = 80k - (2/3)k².
Differentiate: dq/dk = 80 - (4/3)k = 0 → k = 80 × (3/4) = 60 veh/km.
Capacity q_max = 80×60 - (2/3)×60² = 4800 - (2/3)×3600 = 4800 - 2400 = 2400 veh/hr.
Therefore capacity = 2400 veh/hr (option b).
Method-II
Alternatively, recognising the linear form v = vf(1 - k/kj) where vf = 80 km/h and slope (2/3) gives kj = vf / (2/3) = 80 × (3/2) = 120 veh/km. Capacity occurs at k = kj/2 = 60 veh/km, leading to q_max = v(k)×k = 2400 veh/hr.
Method-II
Question 10. 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 [2014 : 2 Marks, Set-II]
(a) 7.5
(b) 12
(c) 40
(d) 60
Ans:(d)
Explanation:
Total travel time = 40 minutes = 40/60 = 2/3 hr. Stopped time t_s = 15 minutes = 0.25 hr. Moving time t_m = 40 - 15 = 25 minutes = 25/60 = 0.416666... hr.
When stopped, the number of vehicles overtaking = q × t_s → q = 45 / 0.25 = 180 veh/hr (stream flow).
When moving, vehicles overtaking per hour (relative) = 60 / t_m = 60 / 0.416666... = 144 veh/hr, which equals q - k V_b where V_b is bicycle speed and k is density.
Bicycle moving speed V_b = distance / moving time = 5 km / (25/60 hr) = 5 / (25/60) = 12 km/hr.
So q - k V_b = 144 → 180 - k×12 = 144 → k×12 = 36 → k = 3 veh/km.
Speed of vehicle stream u = q / k = 180 / 3 = 60 km/hr.
Therefore option (d) 60 km/hr is correct.
Method-I
Method-I
45 vehicles overtook the student when he stopped for 15 minutes.
