## 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?

1. Abhisar

Nuu..

2. Abhisar

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3. Abhisar

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4. Abhisar

Time 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

5. Abhisar

So the answer should be tan -1 <AOB + 90... correct?

6. paki

From where you got these angles...?

7. Abhisar

Which angle?

8. Abhisar

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9. anonymous

You were typing from the morning and you wrote from where you got those angles? @paki :P

10. paki

lml... lag for me... sorry... :/

11. Abhisar

Are you guys able to spot the mistake?

12. paki

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13. anonymous

From where you have taken boat in between this question? You got it for free?

14. Abhisar

@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

15. anonymous

Wait, the man is swimming or he is in the boat??

16. Abhisar

lol, whatever :D I forgot..he is swimming

17. anonymous

See, 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?

18. anonymous

Or relative velocity.. I mean, why you divide that by 2, where is the river speed then?

19. Abhisar

Ummm..not really. Read ex 1 of boat problems here http://www.physicsclassroom.com/class/vectors/Lesson-1/Relative-Velocity-and-Riverboat-Problems

20. paki

Pythagorean theorem

21. anonymous

Okay, I understood, that he will swim regardless or water current, and with his own speed, he will reach to point A..

22. Abhisar

Yes

23. anonymous

*reach point A..

24. anonymous

You have to find resultant velocity buddy..

25. anonymous

Give me some time..

26. Abhisar

ok..

27. anonymous

Okay...

28. anonymous

See, the man has to swim straight, but river flow will make him flow as hypotenuse direction..

29. Abhisar

The 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.

30. anonymous

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31. Abhisar

YEs, man should swim along OB...right?

32. anonymous

Okay then I am misinterpreting the question, wait more then.. :P

33. anonymous

If we move along OB, then river flow must be opposite to reach A. Scratch that diagram.. :)

34. anonymous

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35. anonymous

Okay, let me now again try..

36. Abhisar

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37. anonymous

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38. anonymous

You can use Pythagoras theorem to find the unknown side... $R = \sqrt{2^2 - 1^2 } = \sqrt{3} \quad km/hr$

39. anonymous

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40. Abhisar

one min

41. Abhisar

YaY!!!

42. Abhisar

Done...... dhika chika dhinka chika

43. anonymous

Aise hi hoga na, main to bas tukke jyada aur dimaag kam chala raha tha. :P

44. Abhisar

Pata nahi answer to aa gaya :D

45. anonymous

I 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. :)

46. anonymous

Once you find the unknown, finding angle is just a-jiffy work..

47. Abhisar

Ahaann...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 :)

48. Abhisar

*sleep

49. anonymous

Don't forget to wake up.. :P

50. Abhisar

I wish I could :D