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(C) Let the man start crossing the road at an angle $	heta$ with the perpendicular to the road. The width of the road is $d = 2\,m$. The distance the

Vedclass · Accepted Answer

$3.57$

Vedclass · Answer

(C) Let the man start crossing the road at an angle $	heta$ with the perpendicular to the road. The width of the road is $d = 2\,m$. The distance the man needs to travel to cross the road is $d/\cos	heta = 2/\cos	heta$. The time taken by the man is $t = (2/\cos	heta) / v$.In this time $t$,the truck travels a distance $x = v_0 t = 8 	imes (2 / (v \cos	heta)) = 16 / (v \cos	heta)$.For the man to cross safely,the truck must have passed the point where the man reaches the other side. The initial distance of the truck from the man's path is $4\,m$. The man reaches the other side at a horizontal distance of $d 	an	heta = 2 	an	heta$ from his starting point.Thus,the truck must cover a total distance of $4 + 2 	an	heta$. Therefore,$v_0 t = 4 + 2 	an	heta$.Substituting $t = 2 / (v \cos	heta)$,we get:$8 	imes (2 / (v \cos	heta)) = 4 + 2 	an	heta$$16 / (v \cos	heta) = 4 + 2 (\sin	heta / \cos	heta)$$16 / v = 4 \cos	heta + 2 \sin	heta$$v = 16 / (4 \cos	heta + 2 \sin	heta) = 8 / (2 \cos	heta + \sin	heta)$.To minimize $v$,we maximize the denominator $f(	heta) = 2 \cos	heta + \sin	heta$.$f'(	heta) = -2 \sin	heta + \cos	heta = 0 \implies 	an	heta = 0.5$.For $	an	heta = 0.5$,$\sin	heta = 1/\sqrt{5}$ and $\cos	heta = 2/\sqrt{5}$.$v_{\min} = 8 / (2(2/\sqrt{5}) + 1/\sqrt{5}) = 8 / (5/\sqrt{5}) = 8 / \sqrt{5} \approx 3.57\,m/s$.

$A$ $2\,m$ wide truck is moving with a uniform speed $v_0=8\,m/s$ along a straight horizontal road. $A$ pedestrian starts to cross the road with a uniform speed $v$ when the truck is $4\,m$ away from him. The minimum value of $v$ so that he can cross the road safely is $...........\,m/s$.

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