A person can swim in still water at the rate of 1km/hr. He tries to cross river by swimming perpendicular to the river flowing at the rate of 2km/hr. if the width of the river is 10m find the location of the point where he lands on the other side of the river , also find time taken by him to cross the river. (Answer Fast)

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Solution

Width of the river = 10 m = 0.01 km
Velocity of the river = 2 km/h
Velocity of the man = 1 km/h
Location of the point other side the river , x = 0.01 × 2 / 1 = 0.02 km = 20 m

Time taken to cross the river ,
$t = \frac{\sqrt{s^{2} + x^{2}}}{\sqrt{{v_{2}}^{2} + {v_{1}}^{2}}} = \sqrt{\frac{0 . 01^{2} + 0 . 02^{2}}{2^{2} + 1}} = \sqrt{\frac{5 \times 10^{- 4}}{5}} = 10^{- 2} = 0.01 h r$

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Width of the river = 10 m = 0.01 km Velocity of the river = 2 km/h Velocity of the man = 1 km/h Location of the point other side the river , x = 0.01 × 2 / 1 = 0.02 km = 20 m Time taken to cross the river , t=s2+x2v22+v12=0.012+0.02222+1=5×10-45=10-2 = 0.01 hr