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 =80 m / min vm=(5/3) vr vr2=t (5/3)(vr)2-(80)2 (16/9) vr2=(80)2 vr=60 m / min u2, u0 form canours linear

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

$v_{r}^{2} = v_{m}^{2} - v^{2}$
$v = \frac{320}{4} m / min = 80 m / min$
$v_{m} = \frac{5}{3} v_{r}$
$v_{r}^{2} = t \frac{5}{3} (v_{r})^{2} - (80)^{2}$
$\frac{16}{9} v_{r}^{2} = (80)^{2}$
$v_{r} = 60 m / min$
$u^{2}, u_{0}$ form canours linear

Q. A man crosses a 320m wide river perpendicular to the current in 4min. 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​m/min=80m/min vm​=35​vr​ vr2​=t35​(vr​)2−(80)2 916​vr2​=(80)2 vr​=60m/min u2,u0​ form canours linear

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

$v_{r}^{2} = v_{m}^{2} - v^{2}$
$v = \frac{320}{4} m / min = 80 m / min$
$v_{m} = \frac{5}{3} v_{r}$
$v_{r}^{2} = t \frac{5}{3} (v_{r})^{2} - (80)^{2}$
$\frac{16}{9} v_{r}^{2} = (80)^{2}$
$v_{r} = 60 m / min$
$u^{2}, u_{0}$ form canours linear