The design speed on a road is 60 kmph. Assuming the driver reaction time of 2.5 seconds and coefficient of friction of pavement surface as 0.35, the required stopping distance for two-way traffic on a single lane road is

82.1 m

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102.4 m

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164.2 m

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186.4 m

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Solution

The correct option is C 164.2 m
Stopping Distance
$= 0.278 V t + \frac{V^{2}}{254 f}$

$= 0.278 \times 60 \times 2.5 + \frac{(60)^{2}}{254 \times 0.35} = 82.1 m$

But the traffic is two way, therefore the stopping distance
$= 2 \times 82.1 = 164.2 m$

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The design speed on a road is 60 kmph. Assuming the driver reaction time of 2.5 seconds and coefficient of friction of pavement surface as 0.35, the required stopping distance for two-way traffic on a single lane road is

The correct option is C 164.2 mStopping Distance =0.278Vt+V2254f =0.278×60×2.5+(60)2254×0.35=82.1 m But the traffic is two way, therefore the stopping distance =2×82.1=164.2 m

The correct option is C 164.2 m
Stopping Distance
$= 0.278 V t + \frac{V^{2}}{254 f}$

$= 0.278 \times 60 \times 2.5 + \frac{(60)^{2}}{254 \times 0.35} = 82.1 m$

But the traffic is two way, therefore the stopping distance
$= 2 \times 82.1 = 164.2 m$