Question 1 |
The lane configuration with lane volumes in vehicles per hour of a four-arm
signalized intersection is shown in the figure. There are only two phases: the first
phase is for the East-West and the West-East through movements, and the second
phase is for the North-South and the South-North through movements. There are
no turning movements. Assume that the saturation flow is 1800 vehicles per hour
per lane for each lane and the total lost time for the first and the second phases
together is 9 seconds.
The optimum cycle length (in seconds), as per the Webster's method, is ____________. (round off to the nearest integer)
The optimum cycle length (in seconds), as per the Webster's method, is ____________. (round off to the nearest integer)
12 | |
45 | |
37 | |
65 |
Question 1 Explanation:
From figure
\begin{aligned} y_{N-S}&=\left [ \frac{360}{1800},\frac{396}{1800} \right ]_{max}\\ &=0.22\\ y_{E-W}&=\left [ \frac{504+504}{2 \times 18},\frac{440}{1800} ,\frac{460}{1800}\right ]_{max}\\ &=0.28\\ &\text{Optimum cycle length}\\ &=\frac{1.5L+5}{1-y}\\ &=\frac{1.5 \times 9+5}{1-(y_{N-S}+y_{E-W})}\\ &=\frac{1.5 \times 9+5}{1-(0.22+0.28)}\\ C_o&=37secs. \end{aligned}
\begin{aligned} y_{N-S}&=\left [ \frac{360}{1800},\frac{396}{1800} \right ]_{max}\\ &=0.22\\ y_{E-W}&=\left [ \frac{504+504}{2 \times 18},\frac{440}{1800} ,\frac{460}{1800}\right ]_{max}\\ &=0.28\\ &\text{Optimum cycle length}\\ &=\frac{1.5L+5}{1-y}\\ &=\frac{1.5 \times 9+5}{1-(y_{N-S}+y_{E-W})}\\ &=\frac{1.5 \times 9+5}{1-(0.22+0.28)}\\ C_o&=37secs. \end{aligned}
Question 2 |
Assuming that traffic on a highway obeys the Greenshields model, the speed of
a shockwave between two traffic streams (P) and (Q) as shown in the schematic
is _______ kmph. (in integer)
5 | |
10 | |
15 | |
25 |
Question 2 Explanation:
\text{Speed of shock wave}=\frac{\text{Change in flow}}{\text{Change in density}}
\text{Flow}=\text{Speed} \times \text{Density}
\begin{aligned} Density&=\frac{Flow}{speed}\\ &=\frac{q_Q-q_P}{\frac{q_Q}{V_Q}-\frac{q_P}{V_P}}\\ &=\frac{1800-1200}{\frac{1800}{30}-\frac{1200}{60}}\\ &=\frac{600}{60-20}\\ &=15Kmph \end{aligned}
\text{Flow}=\text{Speed} \times \text{Density}
\begin{aligned} Density&=\frac{Flow}{speed}\\ &=\frac{q_Q-q_P}{\frac{q_Q}{V_Q}-\frac{q_P}{V_P}}\\ &=\frac{1800-1200}{\frac{1800}{30}-\frac{1200}{60}}\\ &=\frac{600}{60-20}\\ &=15Kmph \end{aligned}
Question 3 |
A single-lane highway has a traffic density of 40 vehicles/km. If the time-mean
speed and space-mean speed are 40 kmph and 30 kmph, respectively, the
average headway (in seconds) between the vehicles is
3 | |
2.25 | |
8.33 \times 10^{-4} | |
6.25 \times 10^{-4} |
Question 3 Explanation:
Given that,
Traffic density = 40 Veh/Km.
Time mean speed = 40 Kmph
Space mean speed = 30 kmph
Time Headway (sec) = ?
We know that traffic volume
= Density x Space mean speed = 40x30 =1200 Veh hr.
and we also know that
q=\frac{3600}{H_t}
where H_t- average headway (sc)
1200=\frac{3600}{H_t}
H_t=\frac{3600}{1200}=3 secs.
Traffic density = 40 Veh/Km.
Time mean speed = 40 Kmph
Space mean speed = 30 kmph
Time Headway (sec) = ?
We know that traffic volume
= Density x Space mean speed = 40x30 =1200 Veh hr.
and we also know that
q=\frac{3600}{H_t}
where H_t- average headway (sc)
1200=\frac{3600}{H_t}
H_t=\frac{3600}{1200}=3 secs.
Question 4 |
For a traffic stream, v is the space mean speed, k is the density, q is the flow,
v_f is the free flow speed, and k_j is the jam density. Assume that the speed
decreases linearly with density.
Which of the following relation(s) is/are correct?
Which of the following relation(s) is/are correct?
q=k_jk-\left ( \frac{k_j}{v_f} \right )k^2 | |
q=v_fk-\left ( \frac{v_f}{k_j} \right )k^2 | |
q=v_fv-\left ( \frac{v_f}{k_j} \right )v^2 | |
q=k_jv-\left ( \frac{k_j}{v_f} \right )v^2 |
Question 4 Explanation:
As per Greenshield's
\begin{aligned} V&=V_f\left ( 1-\frac{K}{K_j} \right )\\ q&=K \times V \\ &= K \left [ 1-\frac{K}{K_j} \right ]V_f\\ q&=V_fK-\left ( \frac{V_f}{K_j} \right )K^2\\ V&=V_f\left [ 1-\frac{K}{K_j} \right ] \end{aligned}
From above equation we can say that
\begin{aligned} K&=K_j\left ( 1-\frac{V}{V_f} \right )\\ q&=K \times V \\ q&= K_j \left [ 1-\frac{V}{V_f} \right ] \times V\\ q&=K_jV-\left ( \frac{K_j}{V_f} \right )V^2 \end{aligned}
\begin{aligned} V&=V_f\left ( 1-\frac{K}{K_j} \right )\\ q&=K \times V \\ &= K \left [ 1-\frac{K}{K_j} \right ]V_f\\ q&=V_fK-\left ( \frac{V_f}{K_j} \right )K^2\\ V&=V_f\left [ 1-\frac{K}{K_j} \right ] \end{aligned}
From above equation we can say that
\begin{aligned} K&=K_j\left ( 1-\frac{V}{V_f} \right )\\ q&=K \times V \\ q&= K_j \left [ 1-\frac{V}{V_f} \right ] \times V\\ q&=K_jV-\left ( \frac{K_j}{V_f} \right )V^2 \end{aligned}
Question 5 |
For the dual-wheel carrying assembly shown in the figure, P is the load on each
wheel, a is the radius of the contact area of the wheel, s is the spacing between
the wheels, and d is the clear distance between the wheels. Assuming that the
ground is an elastic, homogeneous, and isotropic half space, the ratio of
Equivalent Single Wheel Load (ESWL) at depth z=d/2 to the ESWL at depth
z=2s is ___________. (round off to one decimal place)
(Consider the influence angle to be 45^{\circ} for the linear dispersion of stress with
depth)
0.2 | |
0.5 | |
0.8 | |
0.1 |
Question 5 Explanation:
\begin{aligned}
\text{ESWL @ depth }\frac{d}{2}&=P \\
\text{ESWL @ depth }2S&=2P \\
\text{Ratio }\frac{P}{2P}&=0.5 \\
\end{aligned}
Question 6 |
At a traffic intersection, cars and buses arrive randomly according to independent
Poisson processes at an average rate of 4 vehicles per hour and 2 vehicles per
hour, respectively. The probability of observing at least 2 vehicles in 30 minutes
is ______. (round off to two decimal places)
0.52 | |
0.64 | |
0.80 | |
0.44 |
Question 6 Explanation:
\begin{aligned}
\lambda _C&=4 \; vehical/hr=2 \; vehical /30 \;min\\
\lambda _C&=2 \; vehical/hr=1 \; vehical /30 \;min\\
\lambda _{Vehical}&=3 \; vehical/30 \;min\\
P(x \lt 2)&=1-P(x\leq 1)\\
&=1-[P(x=0)+P(x=1)]\\
&=1-\left [ \frac{e^{-\lambda }\lambda ^0}{0!} +\frac{e^{-\lambda }\lambda ^1}{1!}\right ]\\
&=1-e^{-3}(1+3)\\
&=1-\frac{4}{e^3}\\
&=0.8
\end{aligned}
Question 7 |
A two-phase signalized intersection is designed with a cycle time of 100 s. The
amber and red times for each phase are 4 s and 50 s, respectively. If the total
lost time per phase due to start-up and clearance is 2 s, the effective green time
of each phase is ______s. (in integer)
96 | |
56 | |
82 | |
48 |
Question 7 Explanation:
Two phase signalized intersections,
Cycle time = 100 secs.
Amber time = 4 secs. [each phase]
Red time = 50 secs. [each phase]
Total lost time = 2 sec. [each phase]
Effective green time = ? [each phase]
Total Effective green = Cycle time - Lost time = 100-2x2 = 96 sec.
As there is symmetry in phases so effective green time per phase= 96/2=48 sec.
Cycle time = 100 secs.
Amber time = 4 secs. [each phase]
Red time = 50 secs. [each phase]
Total lost time = 2 sec. [each phase]
Effective green time = ? [each phase]
Total Effective green = Cycle time - Lost time = 100-2x2 = 96 sec.
As there is symmetry in phases so effective green time per phase= 96/2=48 sec.
Question 8 |
Consider the four points P, Q, R, and S shown in the Greenshields fundamental
speed-flow diagram. Denote their corresponding traffic densities by
k_P,k_Q,k_R and
k_S respectively. The correct order of these densities is
k_P \gt k_Q \gt k_R \gt k_S | |
k_S \gt k_R \gt k_Q \gt k_P | |
k_Q \gt k_R \gt k_S \gt k_P | |
k_Q \gt k_R \gt k_P \gt k_S |
Question 8 Explanation:
Question 9 |
In a three-phase signal system design for a four-leg intersection, the critical flow ratios for each phase are 0.18, 0.32, and 0.22. The total loss time in each of the phases is 2s. As per Webster's formula, the optimal cycle length (in s, round off to the nearest integer) is _______________
28 | |
46 | |
50 | |
58 |
Question 9 Explanation:
\begin{aligned} L &=2 \times 3=6 \text { seconds } \\ n &=3 \\ y &=(0.18+0.32+0.22) \\ C_{0} &=\left(\frac{1.5 L+5}{1-y}\right)=\left(\frac{1.5 \times 6+5}{1-(0.18+0.32+0.22)}\right) \\ &=50 \text { seconds } \end{aligned}
Question 10 |
Relationship between traffic speed and density is described using a negatively sloped straight line. If v_{f} is the free flow speed then the speed at which the maximum flow occurs is
0 | |
\frac{v_{f}}{4} | |
\frac{v_{f}}{2} | |
v_{f} |
Question 10 Explanation:
Speed at maximum flow =\frac{V_{f}}{2}
There are 10 questions to complete.