A man crosses a 320 m wide river perpendicular to the current in 4 min. If in still water he can swim with a speed 5/3 times that of the current, then the speed of the current, in mm-1 in is

Question

A man crosses a 320 m wide river perpendicular to the current in 4 min. If in still water he can swim with a speed 5/3 times that of the current, then the speed of the current, in mm-1 in is (A) 30 (B) 40 (C) 50 (D) 60

Tardigrade Admin · Accepted Answer

vr2=vm2-v2 v=(320/4) m min -1=80 m min -1 vm=(5/3) vr vr2=[(5/3) vr]2-(80)2 ⇒ (16/9) vr2=(80)2 vr=60 m min -1

Q. A man crosses a $320 m$ wide river perpendicular to the current in $4$ min. If in still water he can swim with a speed $5/3$ times that of the current, then the speed of the current, in $m m^{- 1}$ in is

30

40

50

60

$v_{r}^{2} = v_{m}^{2} - v^{2}$
$v = \frac{320}{4} m min^{- 1} = 80 m min^{- 1}$
$v_{m} = \frac{5}{3} v_{r}$
$v_{r}^{2} = [\frac{5}{3} v_{r}]^{2} - (80)^{2}$
$\Rightarrow \frac{16}{9} v_{r}^{2} = (80)^{2}$
$v_{r} = 60 m min^{- 1}$

Q. A man crosses a 320m wide river perpendicular to the current in 4 min. If in still water he can swim with a speed 5/3 times that of the current, then the speed of the current, in mm−1 in is

30

40

50

60

vr2​=vm2​−v2 v=4320​mmin−1=80mmin−1 vm​=35​vr​ vr2​=[35​vr​]2−(80)2 ⇒916​vr2​=(80)2 vr​=60mmin−1

Q. A man crosses a $320 m$ wide river perpendicular to the current in $4$ min. If in still water he can swim with a speed $5/3$ times that of the current, then the speed of the current, in $m m^{- 1}$ in is

$v_{r}^{2} = v_{m}^{2} - v^{2}$
$v = \frac{320}{4} m min^{- 1} = 80 m min^{- 1}$
$v_{m} = \frac{5}{3} v_{r}$
$v_{r}^{2} = [\frac{5}{3} v_{r}]^{2} - (80)^{2}$
$\Rightarrow \frac{16}{9} v_{r}^{2} = (80)^{2}$
$v_{r} = 60 m min^{- 1}$