Question 1 |

The figure presents the time-space diagram for when the traffic on a highway is suddenly stopped for a certain time and then released. Which of the following statements are true?

Speed is higher in Region \mathrm{R} than in Region \mathrm{P} | |

Volume is lower in Region Q than in Region P | |

Volume is higher in Region \mathrm{R} than in Region \mathrm{P} | |

Density is higher in Region \mathrm{Q} than in Region \mathrm{R} |

Question 1 Explanation:

Any vertical line drawn in the graph which cuts arrow lines give positions of vehicles in different regions at a particular time.

Slope of a line in a region represents speed of vehicles in that particular region.

From the Plot :

1. Distance between position of vehicles in region ' P ' is greater than that in region R \Rightarrow Density in region ' P ' is greater than that in region ' R '.

2. Slope of lines in region ' Q ' =0 \Rightarrow Speed of vehicles in region ' \mathrm{Q} ' is zero.

3. Slope of lines in region ' P ' is greater than in region ' R ' \Rightarrow speed of vehicles in region ' P ' is greater than in region ' R '.

4. Since vehicles in region ' Q ' are stationary, flow or volume in region ' Q ' is zero and hence lower than in region \mathrm{P} or region \mathrm{R}. And density in region ' \mathrm{Q} ' is higher than in region \mathrm{P} or region \mathrm{R}.

5. Boundary between region ' P and ' R ' represents 'forward moving slope wave' means slope is positive.

i.e. \quad slope =\frac{q_{R}-q_{P}}{K_{R}-K_{P}} \gt 0

q - Flow

\mathrm{K} - Density.

Since \mathrm{K}_{\mathrm{R}} \gt \mathrm{K}_{\mathrm{P}}

\Rightarrow \mathrm{q}_{\mathrm{R}} should be greater than \mathrm{q}_{\mathrm{P}}.

\Rightarrow Volume/flow in region ' R ' is higher than in Region 'P'.

Slope of a line in a region represents speed of vehicles in that particular region.

From the Plot :

1. Distance between position of vehicles in region ' P ' is greater than that in region R \Rightarrow Density in region ' P ' is greater than that in region ' R '.

2. Slope of lines in region ' Q ' =0 \Rightarrow Speed of vehicles in region ' \mathrm{Q} ' is zero.

3. Slope of lines in region ' P ' is greater than in region ' R ' \Rightarrow speed of vehicles in region ' P ' is greater than in region ' R '.

4. Since vehicles in region ' Q ' are stationary, flow or volume in region ' Q ' is zero and hence lower than in region \mathrm{P} or region \mathrm{R}. And density in region ' \mathrm{Q} ' is higher than in region \mathrm{P} or region \mathrm{R}.

5. Boundary between region ' P and ' R ' represents 'forward moving slope wave' means slope is positive.

i.e. \quad slope =\frac{q_{R}-q_{P}}{K_{R}-K_{P}} \gt 0

q - Flow

\mathrm{K} - Density.

Since \mathrm{K}_{\mathrm{R}} \gt \mathrm{K}_{\mathrm{P}}

\Rightarrow \mathrm{q}_{\mathrm{R}} should be greater than \mathrm{q}_{\mathrm{P}}.

\Rightarrow Volume/flow in region ' R ' is higher than in Region 'P'.

Question 2 |

A plot of speed-density relationship (linear) of two roads (Road A and Road B ) is shown in the figure.

If the capacity of Road A is C_{A} and the capacity of Road B is C_{B}, what is \frac{C_{A}}{C_{B}} ?

If the capacity of Road A is C_{A} and the capacity of Road B is C_{B}, what is \frac{C_{A}}{C_{B}} ?

\frac{k_A}{k_B} | |

\frac{u_A}{u_B} | |

\frac{k_Au_A}{k_Bu_B} | |

\frac{k_Au_B}{k_Bu_A} |

Question 2 Explanation:

For, Road 'A':

\begin{aligned} \mathrm{v}_{\mathrm{f}} & =\mathrm{u}_{\mathrm{A}}, \quad \mathrm{k}_{\mathrm{j}}=\mathrm{k}_{\mathrm{A}} \\ \mathrm{c}_{\mathrm{A}} & =\frac{\mathrm{u}_{\mathrm{A}} \mathrm{k}_{\mathrm{A}}}{4} \\ \mathrm{c}_{\mathrm{A}} & =\text { Capacity of road, } \end{aligned}

For Road 'B':

\begin{aligned} & v_{f}=\text { Free flow speed }=u_{B} \\ & k_{j}=\text { Jam density }=k_{B} \end{aligned}

Capacity of Road 'B' \left(c_{B}\right)=\frac{u_{B} k_{B}}{4}

\therefore \quad \frac{\mathrm{C}_{\mathrm{A}}}{\mathrm{c}_{\mathrm{B}}}=\frac{\mathrm{u}_{\mathrm{A}} \mathrm{k}_{\mathrm{A}}}{4\left(\frac{\mathrm{u}_{\mathrm{B}} \mathrm{k}_{\mathrm{B}}}{4}\right)}=\frac{\mathrm{u}_{\mathrm{A}} \mathrm{k}_{\mathrm{A}}}{\mathrm{u}_{\mathrm{B}} \mathrm{k}_{\mathrm{B}}}

\begin{aligned} \mathrm{v}_{\mathrm{f}} & =\mathrm{u}_{\mathrm{A}}, \quad \mathrm{k}_{\mathrm{j}}=\mathrm{k}_{\mathrm{A}} \\ \mathrm{c}_{\mathrm{A}} & =\frac{\mathrm{u}_{\mathrm{A}} \mathrm{k}_{\mathrm{A}}}{4} \\ \mathrm{c}_{\mathrm{A}} & =\text { Capacity of road, } \end{aligned}

For Road 'B':

\begin{aligned} & v_{f}=\text { Free flow speed }=u_{B} \\ & k_{j}=\text { Jam density }=k_{B} \end{aligned}

Capacity of Road 'B' \left(c_{B}\right)=\frac{u_{B} k_{B}}{4}

\therefore \quad \frac{\mathrm{C}_{\mathrm{A}}}{\mathrm{c}_{\mathrm{B}}}=\frac{\mathrm{u}_{\mathrm{A}} \mathrm{k}_{\mathrm{A}}}{4\left(\frac{\mathrm{u}_{\mathrm{B}} \mathrm{k}_{\mathrm{B}}}{4}\right)}=\frac{\mathrm{u}_{\mathrm{A}} \mathrm{k}_{\mathrm{A}}}{\mathrm{u}_{\mathrm{B}} \mathrm{k}_{\mathrm{B}}}

Question 3 |

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 3 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 4 |

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 4 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 5 |

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 5 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.

There are 5 questions to complete.