Question 1 |
The stopping sight distance (SSD) for a level highway is 140 m for the design speed of 90 km/h. The acceleration due to gravity and deceleration rate are 9.81 \mathrm{~m} / \mathrm{s^2} and 3.5 \mathrm{~m} / \mathrm{s^2}, respectively. The perception/reaction time (in s,round off to two decimal places) used in the SSD calculation is ______________
6.14 | |
5.02 | |
2.02 | |
4.12 |
Question 1 Explanation:
\begin{aligned} \mathrm{SSD} &=140 \mathrm{~m} \\ V &=90 \mathrm{kmph} \\ a &=3.5 \mathrm{~m} / \mathrm{s}^{2} \\ \mathrm{SSD} &=V t_{R}+\frac{V^{2}}{2 g f} \\ a &=g f \\ 140 &=\left(\frac{5}{18} \times 90 \times t_{R}\right)+\frac{\left(\frac{5}{18} \times 90\right)^{2}}{2 \times 3.5} \\ t_{R} &=2.028 \mathrm{~seconds} \end{aligned}
Question 2 |
On a road, the speed - density relationship of a traffic stream is given by u=70 - 0.7k (where speed, u, is in km/h and density, k is in veh/km). At the capacity condition, the average time headway will be
0.5s | |
1.0s | |
1.6s | |
2.1s |
Question 2 Explanation:
\begin{aligned} \mathrm{u}&=70-0.07 \mathrm{k} \\ \mathrm{u}&=70\left[1-\frac{\mathrm{k}}{\frac{70}{0.7}}\right]\\ V_{f} &=70 \mathrm{kmph} \\ \mathrm{k}_{\mathrm{j}} &=\frac{70}{0.7}=100 \mathrm{veh} / \mathrm{km} \\ \mathrm{q}_{\max } &=\frac{1}{4} V_{f} k_{j}=\left(\frac{1}{4} \times 70 \times 100\right)=1750 \mathrm{veh} / \mathrm{hr} \\ \mathrm{q}_{\max } &=\frac{3600}{h_{i}} \\ 1750 &=\frac{300}{h_{i}}\\ & \text{Average time headway}\\ h_{i}&=\frac{3600}{1750}=2.057=2.1 \mathrm{sec} \end{aligned}
Question 3 |
A highway designed for 80 km/h speed has a horizontal curve section with radius 250 m. If the design lateral friction is assumed to develop fully, the required super elevation is
0.02 | |
0.05 | |
0.07 | |
0.09 |
Question 3 Explanation:
\begin{aligned} V &=80 \mathrm{kmph}, R=250 \mathrm{~m} \\ e+f &=\frac{V^{2}}{127 R} \\ e+0.15 &=\frac{80^{2}}{127 \times 250} \\ e &=0.051 \end{aligned}
Question 4 |
The shape of the most commonly designed highway vertical curve is
circular (single radius) | |
circular (multiple radii) | |
parabolic | |
spiral |
Question 4 Explanation:
The ideal vertical curve is a 2^{\circ} parabola.
Question 5 |
The design speed of a two-lane two-way road is 60 km/h and the longitudinal coefficient
of friction is 0.36. The reaction time of a driver is 2.5 seconds. Consider acceleration
due to gravity as 9.8 m/s^2. The intermediate sight distance (in m, round off to the nearest
integer) required for the load is ________
124 | |
246 | |
186 | |
162 |
Question 5 Explanation:
Given : f = 0.36; v = 60 km; g = 9.8 m/s^2;\; t_R=2.5s
\begin{aligned} SSD&= \left ( 0.278Vt_R+\frac{V^2}{254f} \right )\\ &= 0.278 \times 60 \times 2.5 +\frac{60^2}{254 \times 0.36}\\ &= 41.7+39.37=81m\\ ISD &=2 \times SSD\\ &=81 \times 2=162m \end{aligned}
\begin{aligned} SSD&= \left ( 0.278Vt_R+\frac{V^2}{254f} \right )\\ &= 0.278 \times 60 \times 2.5 +\frac{60^2}{254 \times 0.36}\\ &= 41.7+39.37=81m\\ ISD &=2 \times SSD\\ &=81 \times 2=162m \end{aligned}
Question 6 |
In an urban area, a median is provided to separate the opposing streams of traffic. As
per IRC : 86-1983, the desirable minimum width (in m, expressed as integer) of the
median, is __________.
1.2 | |
5 | |
7.5 | |
2.6 |
Question 6 Explanation:
As per IRC : 86-1983
Desirable minimum width of median in urban roads = 5 m
Desirable minimum width of median in urban roads = 5 m
Question 7 |
The coefficient of average rolling friction of a road is f_r and its grade is +G%. If the grade of this road is doubled, what will be the percentage change in the braking distance (for the design vehicle to come to a stop) measured along the horizontal (assume all other parameters are kept unchanged)?
\frac{0.01G}{f_r+0.02G}\times 100 | |
\frac{f_r}{f_r+0.02G}\times 100 | |
\frac{0.02G}{f_r+0.01G}\times 100 | |
\frac{2 f_r}{f_r+0.01G}\times 100 |
Question 7 Explanation:
Percentage change in Braking Distance
\begin{aligned} &=\left ( \frac{L_{B_1}-L_{B_2} }{L_{B_1}} \right ) \times 100\\ \%\Delta L_B&=\frac{\frac{V^2}{254[f_r+0.01G]}-\frac{V^2}{254[f_r+0.02G]}}{\frac{V^2}{254[f_r+0.01G]}}\\ &=\frac{f_r+0.02G-f_r-0.01G}{(f_r+0.01G)(f_r+0.02G)} \times (f_r+0.01G)\\ \%\Delta L_B&=\left ( \frac{0.01G}{f_r+0.02G} \times 100 \right ) \end{aligned}
\begin{aligned} &=\left ( \frac{L_{B_1}-L_{B_2} }{L_{B_1}} \right ) \times 100\\ \%\Delta L_B&=\frac{\frac{V^2}{254[f_r+0.01G]}-\frac{V^2}{254[f_r+0.02G]}}{\frac{V^2}{254[f_r+0.01G]}}\\ &=\frac{f_r+0.02G-f_r-0.01G}{(f_r+0.01G)(f_r+0.02G)} \times (f_r+0.01G)\\ \%\Delta L_B&=\left ( \frac{0.01G}{f_r+0.02G} \times 100 \right ) \end{aligned}
Question 8 |
A 7.5 m wide two-lane road on a plain terrain is to be laid along a horizontal curve of radius 510 m. For a design speed of 100 kmph, super-elevation is provided as per IRC: 73-1980. Consider acceleration due to gravity as 9.81 m/s^{2}. The level difference between the inner and outer edges of the road (in m, up to three decimal places) is _____
0.254 | |
0.525 | |
0.815 | |
1.325 |
Question 8 Explanation:
\begin{aligned} e &=\frac{\sqrt{2}}{225 R}=\frac{100^{2}}{225 \times 510} \\ &=0.0871 \times 0.07=\tan \theta \\ \sin \theta & \simeq 0.07 \\ x &=7.5 \times \sin \theta=0.525 \mathrm{m} \end{aligned}
Question 9 |
A car follows a slow moving truck (travelling at a speed of 10 m/s) on a two-lane two-way highway. The car reduces its speed to 10 m/s and follows the truck maintaining a distance of 16 m from the truck. On finding a clear gap in the opposing traffic stream, the car accelerates at an average rate of 4 m/s^2, overtakes the truck and returns to its original lane. When it returns to its original lane, the distance between the car and the truck is 16 m. The total distance covered by the car during this period (from the time it leaves its lane and subsequently returns to its lane after overtaking) is
64m | |
72m | |
128m | |
144m |
Question 9 Explanation:
Overtaking time,
\begin{array}{l} \quad T=\sqrt{\frac{4 S}{a}}=\sqrt{\frac{4 \times 16}{4}}=4 \mathrm{sec} \\ S=\text { Space heady way }=16 \mathrm{m} \\ a=\text { Acceleration }=4 \mathrm{m} / \mathrm{s}^{2} \end{array}
Distance travelled by vehicle =S_{2}
\begin{aligned} S_{2} &=u T+\frac{1}{2} a T^{2} \\ &=10 \times 4+\frac{1}{2} \times 4 \times 4^{2}=72 \mathrm{m} \end{aligned}
\begin{array}{l} \quad T=\sqrt{\frac{4 S}{a}}=\sqrt{\frac{4 \times 16}{4}}=4 \mathrm{sec} \\ S=\text { Space heady way }=16 \mathrm{m} \\ a=\text { Acceleration }=4 \mathrm{m} / \mathrm{s}^{2} \end{array}
Distance travelled by vehicle =S_{2}
\begin{aligned} S_{2} &=u T+\frac{1}{2} a T^{2} \\ &=10 \times 4+\frac{1}{2} \times 4 \times 4^{2}=72 \mathrm{m} \end{aligned}
Question 10 |
A priority intersection has a single-lane one-way traffic road crossing an undivided two-lane two-way traffic road. The traffic stream speed on the single-lane road is 20 kmph and the speed on the two-lane road is 50 kmph. The perception-reaction time is 2.5s, coefficient of longitudinal friction is 0.38 and acceleration due to gravity is 9.81 m/s^{2}. A clear sight triangle has to be ensured at this intersection. The minimum lengths of the sides of the sight triangle along the two-lane road and the single-lane road, respectively will be
50 m and 20 m | |
61 m and 18 m | |
111 m and 15 m | |
122 m and 36 m |
Question 10 Explanation:
\begin{aligned} \mathrm{SSD}_{2} &=0.278 \times V t_{R}+\frac{V^{2}}{254 f} \\ &=0.278 \times 50 \times 2.5+\frac{50^{2}}{254 \times 0.38} \simeq 61 \mathrm{m} \\ \mathrm{SSD}_{1} &=0.278 \times 20 \times 2.5+\frac{20^{2}}{254 \times 0.38} \simeq 18 \mathrm{m} \end{aligned}
As per data given and based on calculation of SSD the answer would be 61 m and 18 m.
But, as per the code IRC : 66 of sight distance calculation at priority intersection, the sight triangle at priority intersection should be formed by measuring a distance of 15 m along minor road and a distance equal to 8 seconds travel at the design speed (50 \times 0.278 \times 8 \simeq 111 m) along major road. Hence answer will be 111 m and 15 m.
There are 10 questions to complete.