A man crosses a 320 m wide river perpendicul | vedclass

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(D) Let $v_r$ be the speed of the river current and $v_m$ be the speed of the man in still water.Given that the man crosses the river perpendicular to

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$60$

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(D) Let $v_r$ be the speed of the river current and $v_m$ be the speed of the man in still water.Given that the man crosses the river perpendicular to the current,his resultant velocity $v_{res}$ is given by $v_{res} = \sqrt{v_m^2 - v_r^2}$.The width of the river is $d = 320 \ m$ and the time taken is $t = 4 \ min$.Thus,the resultant velocity is $v_{res} = \frac{d}{t} = \frac{320}{4} = 80 \ m/min$.We are given $v_m = \frac{5}{3} v_r$.Substituting this into the velocity equation: $80 = \sqrt{(\frac{5}{3} v_r)^2 - v_r^2}$.Squaring both sides: $80^2 = \frac{25}{9} v_r^2 - v_r^2$.$6400 = \frac{16}{9} v_r^2$.$v_r^2 = \frac{6400 	imes 9}{16} = 400 	imes 9 = 3600$.$v_r = \sqrt{3600} = 60 \ m/min$.

$A$ man crosses a $320 \ m$ wide river perpendicular to the current in $4 \ minutes$. If in still water he can swim with a speed $5/3$ times that of the current,then the speed of the current,in $m/min$ is

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