The speed of overtaking and overtaken vehicles on a highway are $85 k m p h$ and $70 k m p h$ respectively. Calculate the overtaking sight-distance needed for two way traffic. Assume the acceleration of the overtaking vehicle as $2.5 k m p h$ per second and the speed of the vehicle in the opposite direction as $85 k m p h$

545.52

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Solution

The correct option is A 545.52

Given

$V_{a} = 85 k m / h r$
$= 23.61 m / s$

$V_{b} = 70 k m / h r$
$= 19.44 m / s$

$a = 2.5 k m / h r / s e c$
$= 2.5 \times \frac{5}{18} = 0.694 m / {s e c}^{2}$

We know that
$OSD = d_{1} + d_{2} + d_{3} . . . . . . (1)$

$d_{1} = v_{b} t$
$Adopt, t = 2.5 s e c$
$d_{1} = 19.44 \times 2 = 48.6 m$

$d_{2} = b + 2 s = v_{b} T + 2 s . . . . . (2)$

We know that
$T = \sqrt{\frac{4 s}{a}}$
$S = 0.7 v_{b} + 6$
$= 0.7 \times 19.44 + 6 = 19.61 m$
$T = \sqrt{\frac{4 \times 19.61}{0.694}} = 10.632 s e c$

From (2)
$d_{2} = 19.44 \times 10.632 + 2 \times 19.61$
$= 245.9 m$

And,
$d_{3} = v T$
$= 23.61 \times 10.632$
$= 251.02 m$

From (1)
$OSD = 48.6 + 245.9 + 251.02$
$OSD = 545.52 m$

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