Q61GATE 2011MCQ

If the jam density is given as $k_j$ and the free flow speed is given as $u_f$ , the maximum flow for a linear traffic speed-density model is given by which of the following options?

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Q62GATE 2011MCQ

The probability that k number of vehicles arrive (i.e. cross a predefined line) in time t is given as $(\lambda t)^k e^{-\lambda t}/K!, \; where \; \lambda$ is the average vehicle arrival rate. What is the probability that the time headway is greater than or equal to time $t_{1}$ ?

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📖 Explanation

Probability of arriving 'n' number of vehicles in time t $P(n)=\frac{(\lambda t)^{n} e^{-\lambda t}}{n !}$ If time headway is greater than or equal to time t, then number of vehicles arrive is zero, $\Rightarrow \quad P(0) \frac{\left(\lambda t_{1}\right)^{o} e^{-\lambda t_{1}}}{o !}=e^{-\lambda t_{1}}$

Q63GATE 2011MCQ

The cumulative arrival and departure curve of one cycle of an approach lane of a signalized intersection is shown in the figure. The cycle time is 50 s and the effective red time is 30 s and the effective green time is 20 s. What is the average delay?

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📖 Explanation

Average delay

\begin{aligned} &=\frac{\left(\begin{array}{c}\text { Area of triangle formed } \\ \text { by arrival curve }\end{array}\right)-\left(\begin{array}{c}\text { Area of triangle formed } \\ \text { by departure curve }\end{array}\right)}{\text { No. of vehicles }} \\ &=\frac{\frac{1}{2} \times 40 \times 50-\frac{1}{2} \times 40 \times 20}{40} \\ &=15 \mathrm{sec} \end{aligned}

Q64GATE 2010MCQ

As per IRC : 67 : 2001, a traffic sign indicating the Speed Limit on a road should be of

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Q65GATE 2009MCQ

A three-phase traffic signal at an intersection is designed for flows shown in the figure below. There are six groups of flows identified by the numbers 1 through 6. Among these 1, 3, 4 and 6 are through flows and 2 and 5 are right turning. Which phasing scheme is not feasible ?

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Q66GATE 2009MCQ

On a specific highway, the speed-density relationship follows the Greenbeg's model $[v=v_{f}ln(k_{j}/k)]$ , where $v_{f}$ and $k_{j}$ are the free flew speed and jam density respectively. When the highway is operating at capacity, the density obtained as per this model is

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📖 Explanation

Given, $\quad v=v_{f} \log _{e}\left(\frac{k_{j}}{k}\right)$ But Capacity,

\begin{aligned} C &=\text { traffic density } \times \text { speed } \\ &=k \times v \\ \Rightarrow \quad C &=v_{t} k \log _{e}\left(\frac{k_{I}}{k}\right) \end{aligned}

For maximum capacity,

\begin{array}{l} \frac{d C}{d k}=0 \\ \Rightarrow \quad v_{f} \times \frac{d}{d k}\left[k \log _{\theta}\left(\frac{k_{1}}{k}\right)\right]=0 \\ \Rightarrow \quad k \frac{d}{d k} \log _{\theta}\left(\frac{k_{1}}{k}\right)+\log _{e}\left(\frac{k_{j}}{k}\right) \frac{d}{d k} k=0 \\ \Rightarrow \quad k \times \frac{k}{k_{j}} \times k_{l} \times \left(-\frac{1}{k^{2}}\right)+\log _{e} \frac{k_{j}}{k}=0 \\ \Rightarrow \quad \log _{e} \frac{k_{j}}{k}=1 \\ \Rightarrow \quad e=\frac{k_{j}}{k} \\ \Rightarrow \quad k=\frac{k_{j}}{e} \end{array}

Q67GATE 2008MCQ

Design parameters for a signalized intersection are shown in the figure below. The green time calculated for major and minor roads are 34 and 18 s, respectively. The critical lane volume on the major road changes to 440 vehicles per hour per lane and the critical lane volume on the minor road remains unchanged. The green time will

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Q68GATE 2008MCQ

A linear relationship is observed between speed and density on a certain section of a highway. The free flow speed is observed to be 80 km per hour and the jam density is estimated as 100 vehicles per km length. Based on the above relationship, the maximum flow expected on this section and the speed at the maximum flow will respectively be

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📖 Explanation

The maximum flow may be calculated as, $q_{\max }=\frac{\text { free flow speed } \times \text { jam density }}{4}$ $=\frac{80 \times 100}{4}=2000$ vehicles per hour Maximum flow occurs when the speed becomes half of the free flow speed i.e.

\begin{aligned} \text { Speed at } q_{\max } &=\frac{\text { free flow speed }}{2} \\ &=\frac{80}{2}=40 \mathrm{km} \text { per hour } \end{aligned}

Q69GATE 2008MCQ

The capacities of " One-way 1.5 m wide sidewalk (persons per hour)" and "One-way 2 lane urban road (PCU per hour, with no frontage access, no standing vehicles and very little cross traffic)" are respectively

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Q70GATE 2008MCQ

The shape of the STOP sign according to IRC:67-2001 is

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Q71GATE 2008MCQ

A roundabout is provided with an average entry width of 8.4 m, width of weaving section as 14 m, and length of the weaving section between channelizing islands as 35 m. The crossing traffic and total on the weaving section are 1000 and 2000 PCU per hour respectively. The nearest rounded capacity of the roundabout (in PCU per hour) is

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📖 Explanation

Width of the weaving section, W = 14 m Average entry width, e = 8.4 m Length of weaving section between channelizing islands, L =35 m Proportion of weaving traffic is_given by, $p =\frac{\text{crossing traffic on the weaving section}}{\text{total traffic on the weaving section}}$ $\Rightarrow \quad P = \frac{1000}{2000}$ $\Rightarrow \quad p=0.5$ The capacity of the roundabout may be given as,

\begin{aligned} Q=& \frac{280 w\left(1+\frac{e}{w}\right)\left(1-\frac{p}{3}\right)}{1+\frac{w}{L}} \\ Q=& \frac{280 \times 14 \times \left(1+\frac{8.4}{14}\right) \times \left(1-\frac{0.5}{3}\right)}{1+\frac{14}{35}} \\ =& 3733.33 \approx 3700 \mathrm{PCU} \text { per hour } \end{aligned}

Q72GATE 2007MCQ

If a two-lane National Highway and a two-lane State Highway intersect at right angles, the number of potential conflict point at the intersection, assuming that both the roads are two-way is

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Q73GATE 2007MCQ

In signal design as per Indian Roads Congress specifications, if the sum of the ratios of normal flows to saturation flow of two directional traffic is 0.50 and the total lost time per cycle is 10 seconds, the optimum cycle length in seconds is

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📖 Explanation

Given $Y=0.50, L=10 \mathrm{sec}$ $\therefore \quad$ Optimum cycle length, $C_{0}=\frac{1.5 L+5}{1-Y}=\frac{1.5 \times 10+5}{1-0.50}=40 \mathrm{sec}$

Q74GATE 2006MCQ

For designing a 2-phase fixed type signal at an intersection having North-South and East-West road where only straight ahead traffic is permitted, the following data is available. Total time lost per cycle is 12 seconds. The cycle lenght (second) as per webster's approach is

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📖 Explanation

For N-S road and E-W road the higher traffic volume will be taken i.e.

\begin{aligned} q_{1} &=1000 \text { and } q_{2}=900 \\ S_{1} &=2500 \text { and } S_{2}=3000 \\ y_{1} &=\frac{q_{1}}{S_{1}}=\frac{1000}{2500}=0: 4\\ \text{and }\quad y_{2}&=\frac{q_{2}}{S_{2}}=\frac{900}{3000}=0.3 \\ \therefore\quad Y&=y_{1}+y_{2}=0.4+0.3=0.7 \\ \end{aligned}

Optimum cycle time,

\begin{aligned} C_{0} &=\frac{1.5 L+5}{1-Y}=\frac{1.5 \times 12+5}{1-0.7} \\ &=76.67 \approx 77 \mathrm{seconds} \end{aligned}

Q75GATE 2006MCQ

Name the traffic survey data which is plotted by means of "Desire lines".

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Q76GATE 2006MCQ

On an urban road, the free mean speed was measured as 70 kmph and the average spacing between the vehicles under jam condition as 7.0 m. The speed-flow-density equation is given by : $U=U_{sf}[1-\frac{k}{k_{j}}]$ and q=Uk Where, U = space-mean speed (kmph); k= density (veh/km); $U_{sf}$ = free mean speed (kmph); $k_j$ = jam density (veh/km); q = flow (veh/hr). The maximum flow (veh/hr) per lane for this condition is equal to

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📖 Explanation

\begin{aligned} \text { Traffic volume }&=\text { Density } \times \text { Speed } \\ q&=U k \\ \Rightarrow \quad q&=U_{s t}\left(1-\frac{k}{k_{j}}\right) k \\ \Rightarrow \quad q&=U_{s f}\left(k-\frac{k^{2}}{k_{j}}\right) \end{aligned}

Jam density,

\begin{aligned} k_{j}&=\frac{1000}{\text { average spacing between vehicles }} \\ k_{j}&=\frac{1000}{7} \text { and } U_{s t}=70 \mathrm{kmph} \end{aligned}

For maximum traffic volume, $\frac{d q}{d k}=0$

\begin{aligned} \Rightarrow \quad U_{s f}\left(1-\frac{2 k}{k_{j}}\right)&=0 &\left[\because U_{sf}\neq 0 \right]\\ \Rightarrow \quad 1-\frac{2 k}{k_{j}}&=0 \\ \Rightarrow \quad k&=\frac{k_{j}}{2} \end{aligned}

$\therefore \quad$ Maximum traffic volume,

\begin{aligned} q_{\max }&=U_{s f}\left(\frac{k_{j}}{2}-\frac{\left(k_{j} / 2\right)^{2}}{k_{j}}\right) \\ &=U_{s t}\left(\frac{k_{j}}{2}-\frac{k_{j}}{4}\right)=U_{s t}\left(\frac{k_{j}}{4}\right) \\ &=70 \times \frac{1000}{7 \times 4}=2500 veh/hr \end{aligned}

Q77GATE 2005MCQ

A transport company operates a scheduled daily truck service between city P and city Q. One way journey time between these two cities is 85 hours. A minimum layover time of 5 hours is to be provided at each city. How many trucks are required to provide this service ?

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📖 Explanation

Layover time =5 hrs. Total journey time for one round trip

\begin{array}{l} \underset{(A \text { to } B)}{85}+\underset{\text{(Layout)}}{5}\underset{\text(B to A)}{85}\\ =175 \text { hrs. } \end{array}

Number of vehicles $=\frac{175}{24}=7.3 \approx 8 \text { trucks }$ (Rounding on a higher side)

Q78GATE 2005MCQ

A single lane unidirectional highway has a design speed of 65 kmph. The perception-brake-reaction time of drivers is 2.5 seconds and the average length of vehicles is 5 m. The coefficient of longitudinal friction of the pavement is 0.4. The capacity of this road in term of vehicles per hour per lane is

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📖 Explanation

Space headway,

\begin{aligned} S &=0.278 \mathrm{V} t+\frac{V^{2}}{254 \mathrm{f}}+L \\ &=0.278 \times 65 \times 2.5+\frac{(65)^{2}}{254 \times 0.4}+5 \\ &=91.8 \mathrm{m}\\ \text{Capacity,}\\ C &=\frac{1000 \mathrm{V}}{S}=\frac{1000 \times 65}{91.8} \\ &=708.37 \approx 710 \end{aligned}

Q79GATE 2004MCQ

The road geometrics in India are designed for the

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📖 Explanation

The.road geometrics in India are designed for: (a) $30^{th}$ highest hourly traffic volume (b) $98^{th}$ percentile speed

Q80GATE 2003MCQ

The speed and delay studies on a defined section of highway are conducted by

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