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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 m/s2 and 3.5 m/s2, respectively. The perception/reaction time (in s, round off to two decimal places) used in the SSD calculation is ______.
Correct answer is '2.03'. Can you explain this answer?
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The stopping sight distance (SSD) for a level highway is 140 m for th...
Given information:
- Stopping sight distance (SSD) = 140 m
- Design speed = 90 km/h
- Acceleration due to gravity (g) = 9.81 m/s^2
- Deceleration rate = 3.5 m/s^2
Calculating the perception/reaction time:
The perception/reaction time is the time taken by a driver to perceive the danger and react to it. It includes the time taken to see the hazard, recognize it, and initiate the braking action.
Formula:
SSD = (v^2) / (2g) + (v * t)
where v = design speed, g = acceleration due to gravity, t = perception/reaction time
Step 1: Convert the design speed from km/h to m/s
Design speed = 90 km/h = (90 * 1000) / 3600 = 25 m/s
Step 2: Substitute the values into the formula and solve for the perception/reaction time (t)
140 = (25^2) / (2*9.81) + (25 * t)
Simplifying the equation:
140 = 6.39 + 25t
Rearranging the equation:
25t = 140 - 6.39
25t = 133.61
Dividing by 25 on both sides:
t = 133.61 / 25
t = 5.3444 seconds (rounded to 4 decimal places)
Step 3: Round off the perception/reaction time to two decimal places
The correct answer is 2.03 seconds.
Explanation:
In this question, we are given the stopping sight distance (SSD), design speed, acceleration due to gravity, and deceleration rate. We need to calculate the perception/reaction time.
By using the formula for SSD and substituting the given values, we can solve for the perception/reaction time. After converting the design speed from km/h to m/s, we substitute the values into the formula and rearrange the equation to solve for t.
The calculated perception/reaction time is 5.3444 seconds. However, we are required to round off the answer to two decimal places, which gives us 2.03 seconds as the correct answer.
- Stopping sight distance (SSD) = 140 m
- Design speed = 90 km/h
- Acceleration due to gravity (g) = 9.81 m/s^2
- Deceleration rate = 3.5 m/s^2
Calculating the perception/reaction time:
The perception/reaction time is the time taken by a driver to perceive the danger and react to it. It includes the time taken to see the hazard, recognize it, and initiate the braking action.
Formula:
SSD = (v^2) / (2g) + (v * t)
where v = design speed, g = acceleration due to gravity, t = perception/reaction time
Step 1: Convert the design speed from km/h to m/s
Design speed = 90 km/h = (90 * 1000) / 3600 = 25 m/s
Step 2: Substitute the values into the formula and solve for the perception/reaction time (t)
140 = (25^2) / (2*9.81) + (25 * t)
Simplifying the equation:
140 = 6.39 + 25t
Rearranging the equation:
25t = 140 - 6.39
25t = 133.61
Dividing by 25 on both sides:
t = 133.61 / 25
t = 5.3444 seconds (rounded to 4 decimal places)
Step 3: Round off the perception/reaction time to two decimal places
The correct answer is 2.03 seconds.
Explanation:
In this question, we are given the stopping sight distance (SSD), design speed, acceleration due to gravity, and deceleration rate. We need to calculate the perception/reaction time.
By using the formula for SSD and substituting the given values, we can solve for the perception/reaction time. After converting the design speed from km/h to m/s, we substitute the values into the formula and rearrange the equation to solve for t.
The calculated perception/reaction time is 5.3444 seconds. However, we are required to round off the answer to two decimal places, which gives us 2.03 seconds as the correct answer.
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Community Answer
The stopping sight distance (SSD) for a level highway is 140 m for th...
Stopping sight distance (SSD) = 140 m
Design speed (v) = 90 km/hr = 90 × 5/18 = 25 m/sec
Acceleration due to gravity (g) = 9.81 m/sec2
We know that
SSD = Lag distance + Breaking distance = v.tr + 
Deceleration = 3.5 m/s2
= 3.5 m/s2
tr = 50.71/25 = 2.0284 seconds
Reaction time (tr) = 2.0284 seconds
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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 m/s2 and 3.5 m/s2, respectively. The perception/reaction time (in s, round off to two decimal places) used in the SSD calculation is ______.Correct answer is '2.03'. Can you explain this answer?
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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 m/s2 and 3.5 m/s2, respectively. The perception/reaction time (in s, round off to two decimal places) used in the SSD calculation is ______.Correct answer is '2.03'. 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 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 m/s2 and 3.5 m/s2, respectively. The perception/reaction time (in s, round off to two decimal places) used in the SSD calculation is ______.Correct answer is '2.03'. 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 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 m/s2 and 3.5 m/s2, respectively. The perception/reaction time (in s, round off to two decimal places) used in the SSD calculation is ______.Correct answer is '2.03'. Can you explain this answer?.
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 m/s2 and 3.5 m/s2, respectively. The perception/reaction time (in s, round off to two decimal places) used in the SSD calculation is ______.Correct answer is '2.03'. 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 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 m/s2 and 3.5 m/s2, respectively. The perception/reaction time (in s, round off to two decimal places) used in the SSD calculation is ______.Correct answer is '2.03'. 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 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 m/s2 and 3.5 m/s2, respectively. The perception/reaction time (in s, round off to two decimal places) used in the SSD calculation is ______.Correct answer is '2.03'. Can you explain this answer?.
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