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Standard XII

Physics

Relative Velocity

A swimmer swi...

Question

A swimmer swims in still water at a speed = $5 k m / h r$ . He enters a $200 m$ wide river, having river flow speed = $4 k m / h r$ at a point A and proceeds to swim at an angle of $127^{0}$ $(sin 37^{*} - 0.6)$ with the river flow direction. Another point B is located directly across A on the other side. The swimmer lands on the river bank at a point C. from which he walks the distance CB with a speed= $3 k m / h r$ . The total time in which he reach from A to B is

5 m i n

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4 m i n

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3 m i n

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N o n e

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Solution

The correct option is B $4 m i n$
$V_{S R} =$ Velocity of swimmer w.r.t river= $5 k m / h r$
$V_{R} =$ Velocity of river w.r.t ground
$V_{S} =$ Velocity of swimmer w.r.t ground
$V_{S} = V_{S R} + V_{R}$
$V_{S} = (V_{R} - V_{S R} sin 37^{0}) ˆ i + V_{S R} cos 37^{0} ˆ j$
Displacement along x-axis=
$d = (V_{R} - V_{S R} sin 37^{0}) (0.05)$
$= [4 - 5 (\frac{3}{5})] (0.05)$
$= 0.05 k m$
$= 50 m$
Total time= $\frac{50 \times 10^{- 3} k m}{3 k m / h r} + 0.05 h r$
$= \frac{0.05}{3} + 0.05$ $= \frac{0.20}{3} = 0.0666 h r$
$= 4 min$

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