2,m પહોળાઈનો ટ્રક સીધા આડા રસ્તા પર v _0=8,m | vedclass

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Question

(c)

Let the man starts crossing the road at an angle $	heta$ as shown in figure. For safe crossing the condition is that the man must cross the ro

Vedclass · Accepted Answer

$3.57$

Vedclass · Answer

(c)

Let the man starts crossing the road at an angle $	heta$ as shown in figure. For safe crossing the condition is that the man must cross the road by the time the truck describes the distance $4+A C$ or $4+2 \cot 	heta$.

$	herefore \frac{4+2 \cot 	heta}{8}=\frac{2 / \sin 	heta}{v}$

or $\quad v=\frac{8}{2 \sin 	heta+\cos 	heta}$

For minimum $v, \frac{d v}{d 	heta}=0$

or $\frac{-8(2 \cos 	heta-\sin 	heta)}{(2 \sin 	heta+\cos 	heta)^2}=0$

or $2 \cos 	heta-\sin 	heta=0$

or $	an 	heta=2$

From equation (i),

$v_{\min }=\frac{8}{2\left(\frac{2}{\sqrt{5}}ight)+\frac{1}{\sqrt{5}}}=\frac{8}{\sqrt{5}}=3.57\,m / s$

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