\[(a)\; 2.62 m/s \quad (b)\;4.6 m/s \quad (c)\;3.57 m/s \quad (d)1.414 m/s \]

Let the man crossing the road at an angle $\theta$. For safe crossing the man must cross the road by the time the truck describes a distance $4+AC$

or $4+2 \cot \theta$

$=>\large\frac{AC+4}{v_0}=\frac{BC}{v}$

$=\large\frac{4+2 \cot \theta}{8}=\frac{2 /\sin \theta}{v}$

$v= \large\frac{8}{2 \sin \theta+\cos \theta}$

for minimum time

$\large\frac{dv}{d\theta}$$=0=>\large\frac{-8(2 \cos \theta- \sin \theta)}{(2 \sin \theta+\cos \theta)^2}$$=0$

$2 \cos \theta-\sin \theta=0$

$\tan \theta=2$

$=>\sin \theta= \large\frac{2}{\sqrt 5}$

$=>\cos \theta= \large\frac{1}{\sqrt 5}$

Therefore $v= \large\frac{8}{2 \times \Large\frac{2}{\sqrt 5}-\frac{1}{\sqrt 5}}$

$=\large\frac{8}{\sqrt 5}$

$=3.57 m/s$

Hence c is the correct answer.

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