A swimmer crosses the river along the line making an angle of $45^{\circ}$ with the direction of flow. Velocity of the river water is $5 m/s$ . Swimmer takes 6 seconds to cross the river of width $60 m$ . The magnitude of velocity of the swimmer with respect to river will be

10 m/s

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12 m/s

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5 \sqrt{5} m/s

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10 \sqrt{5} m/s

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Solution

The correct option is C $5 \sqrt{5} m/s$
Let
${\to V}_{r, g} =$ Veloctiy of river w.r.t ground
${\to V}_{m, g} =$ Veloctiy of swimmer w.r.t ground
${\to V}_{m, r} =$ Veloctiy of swimmer w.r.t river

Vertical component of ${\to V}_{m, g}$ , $V_{y} = \frac{Distance}{Time} = \frac{60}{6} = 10 m/s$
Since ${\to V}_{m, g}$ makes an angle of $45^{\circ}$ with the direction of river, its horizontal and vertical component will be same.
$∴ {\to V}_{m, g} = 10^i + 10^j$
${\to V}_{r, g} = 5^i$
Relative velocity of swimmer w.r.t river,
${\to V}_{m, r} = {\to V}_{m, g} - {\to V}_{r, g}$
${\to V}_{m, r} = 10^i + 10^j - 5^i = 5^i + 10^j$
Magnitude, $| {\to V}_{m, r} | = \sqrt{5^{2} + 10^{2}} = 5 \sqrt{5} m/s$

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