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The stopping sight distance of a vehicle moving with a speed of 50 kmph in a two lane road is _______ m if the reaction time of driver is 2.3 sec and coefficient of longitudinal friction is 0.38.
  • a)
    57.87
  • b)
    35.65
  • c)
    48.32
  • d)
    12.43
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The stopping sight distance of a vehicle moving with a speed of 50 kmp...
Given data:
- Speed of the vehicle (v) = 50 km/h
- Reaction time of the driver (t) = 2.3 sec
- Coefficient of longitudinal friction (μ) = 0.38

Formula:
The stopping sight distance (SSD) can be calculated using the following equation:
SSD = (v × t) + (v² / (2gμ))

Where,
v = speed of the vehicle
t = reaction time of the driver
g = acceleration due to gravity (9.81 m/s²)
μ = coefficient of longitudinal friction

Calculation:
1. Convert the speed from km/h to m/s:
Speed (v) = 50 km/h = (50 × 1000) / (60 × 60) = 13.89 m/s

2. Calculate the stopping sight distance (SSD) using the formula:
SSD = (v × t) + (v² / (2gμ))
= (13.89 × 2.3) + ((13.89)² / (2 × 9.81 × 0.38))
= 31.95 + (192.72 / 7.44)
= 31.95 + 25.92
= 57.87 m

Therefore, the stopping sight distance of a vehicle moving with a speed of 50 km/h in a two-lane road is 57.87 m. Hence, option (a) is the correct answer.
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Community Answer
The stopping sight distance of a vehicle moving with a speed of 50 kmp...
Stopping sight distance is given by:
= Lag distance + Breaking distance

Calculation:
SSD = 95 m
V = 50 kmph
tR = 2.3 seconds

SSD = 57.87
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The stopping sight distance of a vehicle moving with a speed of 50 kmph in a two lane road is _______ m if the reaction time of driver is 2.3 sec and coefficient of longitudinal friction is 0.38.a)57.87b)35.65c)48.32d)12.43Correct answer is option 'A'. Can you explain this answer?
Question Description
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