For a highway with design speed of 100 kmph, the safe overtaking sight distance is (assume acceleration as $0.53 m / s^{2}$ )

300 m

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750 m

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320 m

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470 m

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Solution

The correct option is B 750 m

Overtaking sight Distance (OSD)

$= d_{1} + d_{2} + d_{3}$

$V_{b} = (V - 16) = (100 - 16) = 84 k m / h r$

$d_{1} = 0.278 V_{b} t$

$= 0.278 \times 84 \times 2.5 = 58.38 m$

$d_{2} = 0.278 V_{b} T + 2 S$

$and T = \sqrt{\frac{4 S}{a}}$

also,
$S = 0.2 V_{b} + 6$

$= 0.2 \times 84 + 6 = 22.8 m$

$\Rightarrow T = \sqrt{\frac{4 \times 22.8}{0.53}} = 13.118 s e c$

$∴ d_{2} = (0.278 \times 84 \times 13.118) + 2 \times 22.8$

$= 306.331 + 45.60 = 351.931 m$

$d_{3} = 0.278 V_{c} T$

$= 0.278 \times 100 \times 13.118$

$= 364. 68 m$

$∴ OSD = d_{1} + d_{2} + d_{3}$

$= 58.38 + 351.931 + 364.68$

$= 774.991 m$

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For a highway with design speed of 100 kmph, the safe overtaking sight distance is (assume acceleration as 0.53 m/s2)

For a highway with design speed of 100 kmph, the safe overtaking sight distance is (assume acceleration as $0.53 m / s^{2}$ )