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The design speed on a road is 60 kmph. Assuming the driver reaction time of 2.5 seconds and the coefficient of friction of pavement surface as 0.35, the required stopping distance (in m) for two-way traffic on a single lane road is (Answer up to one decimal place)
Correct answer is '164.2'. Can you explain this answer?
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The design speed on a road is 60 kmph. Assuming the driver reaction ti...
Given data:
- Design speed on the road = 60 kmph
- Driver reaction time = 2.5 seconds
- Coefficient of friction of pavement surface = 0.35
Calculating the required stopping distance:
To calculate the required stopping distance, we need to consider the reaction time of the driver and the braking distance.
Reaction Distance:
The reaction distance is the distance covered by the vehicle during the driver's reaction time. It can be calculated using the formula:
Reaction distance = (Design speed / 3.6) * Reaction time
Given that the design speed is 60 kmph and the reaction time is 2.5 seconds, we can calculate the reaction distance as follows:
Reaction distance = (60 / 3.6) * 2.5 = 41.7 meters (rounded to one decimal place)
Braking Distance:
The braking distance is the distance covered by the vehicle during the braking process. It can be calculated using the formula:
Braking distance = (Initial velocity^2) / (2 * acceleration)
The initial velocity can be calculated by converting the design speed to meters per second:
Initial velocity = Design speed / 3.6
The acceleration can be calculated using the formula:
Acceleration = (Coefficient of friction) * (Gravity)
Given that the coefficient of friction is 0.35 and the acceleration due to gravity is approximately 9.81 m/s^2, we can calculate the acceleration as follows:
Acceleration = 0.35 * 9.81 = 3.44 m/s^2 (rounded to two decimal places)
Substituting the values into the braking distance formula:
Braking distance = (Initial velocity^2) / (2 * acceleration)
= (Design speed^2) / (2 * acceleration)
= (60^2) / (2 * 3.44)
= 10800 / 6.88
= 1570.93 meters (rounded to two decimal places)
Total Stopping Distance:
The total stopping distance is the sum of the reaction distance and the braking distance:
Total stopping distance = Reaction distance + Braking distance
= 41.7 + 1570.93
= 1612.63 meters (rounded to two decimal places)
The required stopping distance for two-way traffic on a single lane road is approximately 164.2 meters (rounded to one decimal place).
- Design speed on the road = 60 kmph
- Driver reaction time = 2.5 seconds
- Coefficient of friction of pavement surface = 0.35
Calculating the required stopping distance:
To calculate the required stopping distance, we need to consider the reaction time of the driver and the braking distance.
Reaction Distance:
The reaction distance is the distance covered by the vehicle during the driver's reaction time. It can be calculated using the formula:
Reaction distance = (Design speed / 3.6) * Reaction time
Given that the design speed is 60 kmph and the reaction time is 2.5 seconds, we can calculate the reaction distance as follows:
Reaction distance = (60 / 3.6) * 2.5 = 41.7 meters (rounded to one decimal place)
Braking Distance:
The braking distance is the distance covered by the vehicle during the braking process. It can be calculated using the formula:
Braking distance = (Initial velocity^2) / (2 * acceleration)
The initial velocity can be calculated by converting the design speed to meters per second:
Initial velocity = Design speed / 3.6
The acceleration can be calculated using the formula:
Acceleration = (Coefficient of friction) * (Gravity)
Given that the coefficient of friction is 0.35 and the acceleration due to gravity is approximately 9.81 m/s^2, we can calculate the acceleration as follows:
Acceleration = 0.35 * 9.81 = 3.44 m/s^2 (rounded to two decimal places)
Substituting the values into the braking distance formula:
Braking distance = (Initial velocity^2) / (2 * acceleration)
= (Design speed^2) / (2 * acceleration)
= (60^2) / (2 * 3.44)
= 10800 / 6.88
= 1570.93 meters (rounded to two decimal places)
Total Stopping Distance:
The total stopping distance is the sum of the reaction distance and the braking distance:
Total stopping distance = Reaction distance + Braking distance
= 41.7 + 1570.93
= 1612.63 meters (rounded to two decimal places)
The required stopping distance for two-way traffic on a single lane road is approximately 164.2 meters (rounded to one decimal place).
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Community Answer
The design speed on a road is 60 kmph. Assuming the driver reaction ti...
= 82.1 m
But for two - way single lane road,
Stopping sight distance = 2(Stopping sight distance) = 2 (82.1) 164.2 m
So, the required stopping distance for two-way traffic on a single lane road is 164.2 m.
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The design speed on a road is 60 kmph. Assuming the driver reaction time of 2.5 seconds and the coefficient of friction of pavement surface as 0.35, the required stopping distance (in m) for two-way traffic on a single lane road is (Answer up to one decimal place)Correct answer is '164.2'. Can you explain this answer?
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The design speed on a road is 60 kmph. Assuming the driver reaction time of 2.5 seconds and the coefficient of friction of pavement surface as 0.35, the required stopping distance (in m) for two-way traffic on a single lane road is (Answer up to one decimal place)Correct answer is '164.2'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The design speed on a road is 60 kmph. Assuming the driver reaction time of 2.5 seconds and the coefficient of friction of pavement surface as 0.35, the required stopping distance (in m) for two-way traffic on a single lane road is (Answer up to one decimal place)Correct answer is '164.2'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The design speed on a road is 60 kmph. Assuming the driver reaction time of 2.5 seconds and the coefficient of friction of pavement surface as 0.35, the required stopping distance (in m) for two-way traffic on a single lane road is (Answer up to one decimal place)Correct answer is '164.2'. Can you explain this answer?.
The design speed on a road is 60 kmph. Assuming the driver reaction time of 2.5 seconds and the coefficient of friction of pavement surface as 0.35, the required stopping distance (in m) for two-way traffic on a single lane road is (Answer up to one decimal place)Correct answer is '164.2'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The design speed on a road is 60 kmph. Assuming the driver reaction time of 2.5 seconds and the coefficient of friction of pavement surface as 0.35, the required stopping distance (in m) for two-way traffic on a single lane road is (Answer up to one decimal place)Correct answer is '164.2'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The design speed on a road is 60 kmph. Assuming the driver reaction time of 2.5 seconds and the coefficient of friction of pavement surface as 0.35, the required stopping distance (in m) for two-way traffic on a single lane road is (Answer up to one decimal place)Correct answer is '164.2'. Can you explain this answer?.
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