Road Planning and Sight Distance

JKSSB Junior Engineer (Civil) · Highway Engineering · 11 practice questions with answers and explanations: 6 Easy, 3 Medium, 2 Hard.

Sample questions

Sample Easy question 1

The overtaking sight distance includes the distance travelled during:

  1. The reaction time, the overtaking operation and by the opposing vehicle
  2. Only the braking time
  3. Only the reaction time
  4. The parking of the vehicle
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Answer: A. It is the sum d₁ + d₂ + d₃.

Sample Easy question 2

The speed limit on a road is usually fixed from the:

  1. Maximum speed ever recorded
  2. 85th percentile speed of the vehicles
  3. Slowest vehicle
  4. Average speed of all vehicles
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Answer: B. It is the speed below which 85% of the vehicles travel.

Sample Easy question 3

The design speed of a road is the:

  1. Maximum safe speed on which the geometric design is based
  2. Average speed of traffic
  3. Speed limit shown on signs
  4. Speed of the slowest vehicle
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Answer: A. Sight distance and curve radii are worked out from it.

Sample Medium question 4

The stopping sight distance at 60 km/h with a reaction time of 2.5 s and f = 0.35 is about:

  1. 250 m
  2. 164 m
  3. 41 m
  4. 82 m
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Answer: D. 0.278 × 60 × 2.5 + 60² ÷ (254 × 0.35) = 41.7 + 40.5 = 82.2 m.

Sample Medium question 5

The stopping sight distance on a road is 80 m. The intermediate sight distance is:

  1. 240 m
  2. 160 m
  3. 40 m
  4. 80 m
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Answer: B. Intermediate sight distance is twice the stopping sight distance.

Sample Medium question 6

Vehicles travelling at 60 km/h with an average spacing of 30 m give a lane capacity (q = 1,000 V ÷ S) of:

  1. 1,800 vehicles per hour
  2. 1,000 vehicles per hour
  3. 4,000 vehicles per hour
  4. 2,000 vehicles per hour
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Answer: D. q = 1,000 × 60 ÷ 30 = 2,000 vehicles per hour.

Sample Hard question 7

A car at 60 km/h on a 3% downward gradient, with a reaction time of 2.5 s and f = 0.35, has a stopping sight distance of about:

  1. 86 m
  2. 79 m
  3. 44 m
  4. 82 m
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Answer: A. SSD = 0.278 V t + V² ÷ [254 (f − 0.01 n)] = 41.7 + 44.3 = 86 m.

Sample Hard question 8

A traffic stream has 100 cars (1 PCU each) and 20 trucks (3 PCU each). The flow in PCU is:

  1. 320
  2. 120
  3. 100
  4. 160
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Answer: D. 100 × 1 + 20 × 3 = 160 PCU.

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