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A man can walk on the shore at a speed v...

A man can walk on the shore at a speed `v_1` = 6 km/hr & swim in still water with a speed `V_2` = 5 km/hr. If the speed of water is `V_3` = 4 km/hr, at what angle should he head in the river in order to reach the exactly opposite point of the other shore in "shortest time including his swimming & walking?

Text Solution

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Directly using the previous result we obtain the angle of swimming `theta`-`sin^(-1)(V_(mw))/(V+V_(w))`
Putting v = `V_1` = 6 km/hr, `V_(mw)=V_(2)` = 5 km/hr `V_(w) - V_(3)` = 4 km/hr we obtain`theta `=` sin^(-1)(1/2) = 30^(@)` The man should head at an angle of a = `90^(@)` + 0 = `120^(@)` with the direction of flow of water.
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