The speed of overtaking vehicle is 90 kmph and acceleration is 3.0 kmph/sec. The desirable length of overtaking zone for one way traffic should be

[Assume the length of the vehicle as 6 m and reaction time as 2 sec]

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Solution

Design speed = Speed of overtaking vehicle
$\Rightarrow V = 90 K m p h$

Speed of overtaken vehicle,
$(V_{b}) = V - 16 = 74 k m p h$

Overtaking sight distance = $d_{1} + d_{2}$

Given reaction time is 2 seconds.

$∴ d_{1} = 0.278 \times V_{b} \times t = 0.278 \times 74 \times 2 = 41.144 m$

$d_{2} = 0.278 V_{b} \times T + 2 s$

$S = 0.2 V_{b} + 6 = 0.2 \times 74 + 6 = 20.8 m$

$T = \sqrt{\frac{4 s}{a}};$

$a = 3 k m p h / s e c = (3 \times \frac{5}{18}) m / s^{2}$

$∴ T = \sqrt{\frac{4 \times 20.8}{3 \times \frac{5}{18}}} = 9.99 s e c$

Hence,
$d_{2} = 0.278 \times 74 \times 9.99 + 2 \times 20.8$
$= 247.11 m$

$∴ O S D = d_{1} + d_{2}$
$= 41.144 + 247.11$
$= 288.254 m$

Desirable length of overtaking zone
$= 5 \times O S D$
$= 1441.27 m ≃ 1441.5 m (s a y)$

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