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**Question 1. The speed-density relationship of a highway is given asu = 100 - 0.5kwhere, u = speed in km per hour, k = density in vehicles per km. The maximum flow (in vehicles per hour, round off to the nearest integer) is _____ [2019 : 2 Marks, Set-II]Solution: **

Solution:

Service rate = 50 veh/min

R= 30 sec, G

Time corresponding to which no. of arrival becomes same as no. of departure.

=> 20 x (R + t) =50t

20 x (30 + t) =50t

600 + 20t =50t

Solution:

(a)

(b)

(c)

(d)

Answer:

Solution:

Solution:

V

k

As per linear model (green-shield)

q = vk

For initial values of k, v is constant

∴ q will very linearly with k

Later on, v varies linearly with k.

∴ q will vary parabolically with k.

**Question 13. The safety within a roundabout and the efficiency of a roundabout can be increased, respectively by [2017 : 2 Marks, Set-II](a) increasing the entry radius and increasing the exit radius(b) increasing the entry radius and decreasing the exit radius(c) decreasing the entry radius and increasing the exit radius(d) decreasing the entry radius and decreasing the exit radius Answer: **(c)

At time, t = 0, the light has just turned red. The effective green time is 36 seconds, during which vehicles discharge at the saturation flow rate, s (in vph). Vehicles arrive at a uniform rate, v(in vph), throughout the cycle. Which one of the following statements is TRUE? [2017 : 2 Marks, Set-I]

(a) v = 600 vph, and for this cycle, the average stopped delay per vehicle = 30 seconds.

(b) s = 1800 vph, and for this cycle, the average stopped delay per vehicle = 28.125 seconds.

(c) v = 600 vph, and for this cycle, the average stopped delay per vehicle = 45 seconds.

(d) s = 1200 vph, and for this cycle, the average stopped delay per vehicle = 28.125 seconds.

Answer:

Solution:

Solution:

λ = 900 veh/hour

Probability that time gap (time headway) betwen successive vehicles is greater than 8 seconds,

**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 Answer:** (a)

Solution:

(a) 15

(b) 20

(c) 25

(d) 30

Answer:

Solution:

Let us assume,

q = traffic flow

k = traffic density

V = velocity of test vehicle along the stream

V - velocity of test vehicle against the stream

=> q - KV

=> q - k x 20 = 50 ...(i)

q + kV

=> q + k x 30 = 200 ...(ii)

Solving equation (i) and (ii), we get

k = 3 Veh/km

Consider a ‘L’ km long stretch,

Solution:

= 3.0 sec

Flow of traffic stream

Solution:

q= uk

q= (70-0.7 k)k

= 70k-0.7 k

Maximum possible volume is capacity of road

For maximum volume,

(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)

(a) u

(b) u

(c) u

(d) u

Solution:

q

q

and q

Saturation flow rate is 1800 veh/hr/lane

Lost time,

L = 4 x 4 = 16 sec

Length of the cycle,

C

(a) 1200

(b) 2400

(c) 4800

(d) 9600

45 vehicles overtook the student when he stopped for 15 minutes.

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