For a two-way traffic road, the following are the particulars:
Speed of overtaking vehicle = 65 Kmph
Speed difference between the vehicles = 15 kmph
Acceleration of overtaking vehicle = 3.28 kmph/sec
Perception time of driver of overtaking vehicle = 2 seconds
Length of overtaking vehicle = 6 m
The length of safe overtaking sight distance will be

384 m

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286 m

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328 m

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302 m

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Solution

The correct option is C 328 m
$L = [0.278 V_{2} t + (0.278 V_{2} t_{0} + 2 S) + 0.278 V t_{0}] . . . . . (i)$

In this problem,

$V_{2} = 65 - 15 = 50 k m p h$
$t = 2 s e c$
$V = 65 k m p h$
$S = (0.2 \times 50 + 6) = 16 m$

$t_{0} = \sqrt{\frac{4 \times S}{a}} = \sqrt{\frac{4 \times 16}{3.28 \times \frac{5}{18}}} = 8.38 sec$
Substituting in (i),
$L = [0.278 \times 50 \times 2 + (0.278 \times 50 \times 8.38 + 2 \times 16) + 0.278 \times 65 \times 8.38]$
$= 327.8 m ≃ 328 m$
So, the nearest answer is option (a).

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