A person is ordered to swim across a river of width 500 m. A what minimum rate should he swim perpendicular to river flow so that horizontal drift is 300 m only down the stream. Speed of water current is $3 km {hr}^{- 1} .$

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$5 km {hr}^{- 1}$

$6 km {hr}^{- 1}$

$3 km {hr}^{- 1}$

$4 km {hr}^{- 1}$

answer is C.

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Detailed Solution

Time to cross the river $t = \frac{d}{v_{P}}$ and drift.

$x = v_{R} \times t$

$So x = v_{R} \frac{d}{v_{p}}$

$300 = 3 km {hr}^{- 1} \times \frac{500}{v_{P}}$

$v_{P} = 5 km {hr}^{- 1}$

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