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Abhisar
 one year ago
A man wishes to swim across a river 0.5 m wide. If he can swim at the rate of 2 Km/ h in still water and the river flows at the rate of 1 km/hr. The angle (wrt flow of water) along which he should swim so as to reach a point exactly opposite his starting point, should be?
Abhisar
 one year ago
A man wishes to swim across a river 0.5 m wide. If he can swim at the rate of 2 Km/ h in still water and the river flows at the rate of 1 km/hr. The angle (wrt flow of water) along which he should swim so as to reach a point exactly opposite his starting point, should be?

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Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1Time taken to cross the river = 0.5/2 = 1/4 Now in the same time the boat will travel a horizontal distance of AB. So AB = 1/4 * speed of river water = 0.25 Km Now, AB/BO = tan <AOB

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1So the answer should be tan 1 <AOB + 90... correct?

paki
 one year ago
Best ResponseYou've already chosen the best response.0From where you got these angles...?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You were typing from the morning and you wrote from where you got those angles? @paki :P

paki
 one year ago
Best ResponseYou've already chosen the best response.0lml... lag for me... sorry... :/

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1Are you guys able to spot the mistake?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0From where you have taken boat in between this question? You got it for free?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1@paki the question want us to find the angle with respect to flow of water along which the boat should go. @waterineyes the boat is goingg along the line OA

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait, the man is swimming or he is in the boat??

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1lol, whatever :D I forgot..he is swimming

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0See, his speed of swimming is 2 km/h, and that of river is 1 km/h, he has to swim against the flow, so you need to find relative speed no?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Or relative velocity.. I mean, why you divide that by 2, where is the river speed then?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1Ummm..not really. Read ex 1 of boat problems here http://www.physicsclassroom.com/class/vectors/Lesson1/RelativeVelocityandRiverboatProblems

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, I understood, that he will swim regardless or water current, and with his own speed, he will reach to point A..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You have to find resultant velocity buddy..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0See, the man has to swim straight, but river flow will make him flow as hypotenuse direction..

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1The man wants to reach the exact opposite point so he should swim along the hypotenuse on purpose so that river flow will make him reach on the opp side along a straight line.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437502730218:dw

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1YEs, man should swim along OB...right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay then I am misinterpreting the question, wait more then.. :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If we move along OB, then river flow must be opposite to reach A. Scratch that diagram.. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437502906907:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, let me now again try..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437502994510:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can use Pythagoras theorem to find the unknown side... \[R = \sqrt{2^2  1^2 } = \sqrt{3} \quad km/hr\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437503162276:dw

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1Done...... dhika chika dhinka chika

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Aise hi hoga na, main to bas tukke jyada aur dimaag kam chala raha tha. :P

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1Pata nahi answer to aa gaya :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got its nerve.. See, you have to take river flow and man or boat flow as vectors, draw them on paper to know their directions, on its basis you will get whether to subtract them in squares or to add them. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Once you find the unknown, finding angle is just ajiffy work..

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1Ahaann...I see but there must be another way, something like i was doing earlier. I will mess with it tomorrow. But thank you, cz of uh i can seep in peace now :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Don't forget to wake up.. :P
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