# Highway Sight Distances - IES Previous Year Questions

1.â€ƒWhich one of the following expressions gives intermediate sight distance as per IRC standards ?

**[IES 1999]**

- 2 SSD
- (SSD + OSD) /2
- (OSD – SSD) /2
- 2 OSD

**Answer:**Option A

**Explanation:**

As per IRC standards, Intermediate Sight Distance (ISD) = 2 x Stopping Sight Distance (SSD)

2. â€ƒTotal reaction time of a driver does not depend upon _____ .

**[IES 2000]**

- Perception time
- Brake reaction time
- Condition of mind of the driver
- Speed of vehicle

**Answer:**Option C

**Explanation:**

- Total reaction time is the sum of perception time and brake reaction time.
- The perception time depends on the speed of the vehicle. Hence, the total reaction time also depends on the speed on the vehicle.

3. â€ƒWhile driving at a speed of 30 kmph (with available friction 0.4) down the grade, the driver requires a braking distance twice that required for stopping the vehicle when he travels up the same grade. The grade is _____ .

**[IES 2001]**

- 7%
- 10.6%
- 13.3%
- 33.3%

**Answer:**Option C

**Explanation:**

Braking distance for downward gradient, SSD

_{1}=

^{2}/ 2g(f - 0.01n)

Braking distance for upward gradient, SSD

_{2}=

^{2}/ 2g(f + 0.01n)

Given that, SSD

_{1}= 2 x SSD

_{2}

⇒

^{2}/ 2g(f - 0.01n)

^{2}/ 2g(f + 0.01n)

⇒ f + 0.01n = 2 x (f - 0.01n)

⇒ f = 0.03n

⇒ n = 0.4/0.03

⇒ n = 13.33%

4. â€ƒConsider the following factors:

- Reaction time
- Speed
- Coefficient of longitudinal friction
- Gradient

Which of these factors are taken into account for computing braking distance ?

**[IES 2002]**

- 1 and 3
- 1, 2 and 4
- 2, 3 and 4
- 2 and 3

**Answer:**Option C

**Explanation:**

Braking distance =

^{2}/ 2g(ηf ± 0.01n)

Thus, braking distance depends on speed of the vehicle (v), braking effeciency (η), coefficient of longitudinal friction (f) and gradient (n).

5. â€ƒFor a given road, safe stopping sight distance is 80 m and passing sight distance is 300 m. What is the intermediate sight distance ?

**[IES 2004]**

- 220 m
- 190 m
- 160 m
- 150 m

**Answer:**Option C

**Explanation:**

As per IRC standards, Intermediate Sight Distance (ISD) = 2 x Stopping Sight Distance (SSD)

⇒ ISD = 2 x SSD = 2 x 80

⇒ ISD = 160 m

6. â€ƒWhat will be the non-passing sight distance on a highway for a design speed of 100 kmph when its ascending gradient is 2%, assuming coefficient of friction as 0.7 and brake efficiency as 50% ?

**[IES 2015]**

- 176 m
- 200 m
- 150 m
- 185 m

**Answer:**Option A

**Explanation:**

Design speed, v = 100 kmph = 27.78 m/s

Non-passing Sight Distance for ascending gradient = vt +

^{2}/ 2g(Î·f + 0.01n)

⇒ SD = (27.78 x 2.5) +

^{2}/ 2 x 9.81 x {(0.5 x 0.7) + (0.01 x 2)}

⇒ SD = 175.76 m

⇒ SD ≈ 176 m

7. â€ƒConsider two cars approaching from the opposite directions at 90 kmph and 60 kmph. If the reaction time is 2.5 s, coefficient of friction is 0.7 and brake efficiency is 50% in both the cases, the minimum sight distance required to avoid a head-on collision will be nearly _____ .

**[IES 2019]**

- 188 m
- 212 m
- 236 m
- 154 m

**Answer:**Option C

**Explanation:**

Design speed, v

_{1}= 90 kmph = 25 m/s

Design speed, v

_{2}= 60 kmph = 16.67 m/s

For 90 kmph car, Stopping Sight Distance SSD

_{1}= v

_{1}t +

_{1}

^{2}/ 2g(Î·f)

⇒ SSD

_{1}= (25 x 2.5) +

^{2}/ 2 x 9.81 x (0.5 x 0.7)

⇒ SSD

_{1}= 153. 51 m

For 60 kmph car, Stopping Sight Distance SSD

_{2}= v

_{2}t +

_{2}

^{2}/ 2g(Î·f)

⇒ SSD

_{1}= (16.67 x 2.5) +

^{2}/ 2 x 9.81 x (0.5 x 0.7)

⇒ SSD

_{2}= 82.14 m

Since, the cars are approaching in opposite directions, required Stopping Sight Distance SSD = SSD

_{1}+ SSD

_{2}

⇒ SSD = 153.51 + 82.14

⇒ SSD = 235.65 m ≈ 236 m

8. â€ƒThe sight distance available on a road to a driver at any instance depends on _____ .

- Features of the road ahead.
- Height of driverâ€™s eye above the road surface.
- Height of the object above the road surface.

**[IES 2020]**

- 1 and 2 only
- 1 and 3 only
- 2 and 3 onlyÂ
- 1, 2 and 3

**Answer:**Option D

**Explanation:**

The sight distance depends on all the above mentioned conditions.

9. â€ƒConsider the following data:

Design speed = 96 kmph, Speed of overtaken vehicle = 80 kmph, Reaction time for overtaking = 2 sec, Acceleration = 2.5 kmph/sec.

The safe overtaking sight distance on a two-way traffic road will be nearly _____ .

**[IES 2020]**

- 646 m
- 556 m
- 466 m
- 376 m

**Answer:**Option A

**Explanation:**

Design speed, v = 96 kmph = 26.67 m/s

Speed of overtaken vehicle, v

_{b}= 80 kmph = 22.22 m/s

Acceleration, a = 2.5 kmph/sec = 0.69 m/s

^{2}

Space headway, S = 0.7v

_{b}+ 6

⇒ S = (0.7 x 22.22) + 6 = 21.55 m

Time taken for overtaking, T = √ 4S/a

⇒ T = √ (4 x 21.55)/0.69 = 11.18 s

The safe Overtaking Sight Distance, OSD = d

_{1}+ d

_{2}+ d

_{3}

Here, d

_{1}= v

_{b}x t = 22.22 x 2 = 44.44 m

d2 = b + 2S = (v

_{b}x T) + 2S = (22.22 x 11.18) + (2 x 21.55) = 291.52 m

d3 = vT = 26.67 x 11.18 = 298.17 m

Thus, OSD = 44.44 + 291.52 + 298.17 = 634.13 m

10. â€ƒWhat is the value of headlight sight distance for a highway with a design speed of 65 kmph ? (Take f = 0.36 and t = 2.5 sec)

**[IES 2021]**

- 66.5 m
- 81.3 m
- 91.4 m
- 182.8 m

**Answer:**Option C

**Explanation:**

Design speed, v = 65 kmph = 18.06 m/s

Stopping Sight Distance, SSD = vt +

^{2}/ 2gf

⇒ SSD = (18.06 x 2.5) +

^{2}/ 2 x 9.81 x 0.36

⇒ SSD = 91.33 m

Thus, Headlight Stopping Distance (HSD) = SSD = 91.33 m