A swimmer crosses the river so as to reach the opposite point, along the line making an angle of $45^{o}$ with the direction of flow. Velocity of the river water is $5 m / s$ . Swimmer takes 12 seconds to cross the river of width $60 m$ . The velocity of the swimmer (in $m / s$ ) with respect to flowing water will be:

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Solution

We have, breadth of the river as $B = 60 m$ and time take to cross the river as $t = 12 s$ .
$v_{y} = \frac{B}{t} v_{y} = \frac{60}{12} = 5 m / s$

For $45^{o}$ angle
$v_{x} = v_{y} ∴ v_{x} = 5 m / s_{M} = 5^i + 5^j_{R} = 5^i ∴_{M R} =_{M} -_{R} = 5^j$

$_{M R} = 5 m / s$

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We have, breadth of the river as B=60 m and time take to cross the river as t=12 s. vy=Btvy=6012=5 m/s For 45o angle vx=vy∴vx=5 m/s−→vM=5^i+5^j−→vR=5^i∴−−→vMR=−→vM−−→vR=5^j −−→vMR=5 m/s

We have, breadth of the river as $B = 60 m$ and time take to cross the river as $t = 12 s$ .
$v_{y} = \frac{B}{t} v_{y} = \frac{60}{12} = 5 m / s$

For $45^{o}$ angle
$v_{x} = v_{y} ∴ v_{x} = 5 m / s_{M} = 5^i + 5^j_{R} = 5^i ∴_{M R} =_{M} -_{R} = 5^j$

$_{M R} = 5 m / s$