## Abhisar one year ago A man wishes to swim across a river 0.5 Km 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

@ganeshie8 @nincompoop @dan815

2. Abhisar

|dw:1437535101502:dw|

3. anonymous

Your sketch is accurate. You are after <AOB. You have two sides of a right triangle, OA=2 and AB=1. Trigonometry will give you the answer. You OK with the trig ratios?

4. Abhisar

Yes, I know this way to solve it but I was seeking a different way to do it. Would you mind checking it?

5. anonymous

6. Abhisar

Ok, since the river is 0.5 Km wide and the swimmer can swim with a speed of 2 km/h it will take 0.5/2 = 0.25 hours for him to reach the other side. The swimmer will actually aim along AO but due to the river flow he will reach at B. While reaching B he has travelled a horizontal distance of 1 * 0.25 = 0.25 Km due to river flow. |dw:1437535731457:dw|

7. Abhisar

Now AB/OB = tan <AOB which gives a different answer...

8. Abhisar

@ospreytriple

9. anonymous

He swims at 2 km/hr, but if he is travelling along segment OA, he is not making progress across the river at 2 km/hr, but rather at 2cos<AOB km/hr.

10. Abhisar

Yes but isn't he still making 2km/h across the vertical direction?

11. anonymous

Nope. He is travelling at 2 km/hr at 30 degrees CCW to the vertical

12. anonymous

So he is making progress across the river at 2cos 30 = 1.73 km/hr.

13. Abhisar

Yes, that makes sense now.

14. anonymous

Super. Good question.

15. Abhisar

Thank you.

16. nincompoop

what do you understand about the question

17. nincompoop

not the given, but the question it wants you to solve for.

18. anonymous

You're welcome

19. Abhisar

I already tried it according to my understanding above, turns out i was taking 2 km/h as vertical velocity of swimmer but it is actually his velocity across line Ao

20. nincompoop

|dw:1437536977448:dw|

21. Abhisar

You are doing the same mistake I guess. I thought this way too but actually the swimmer will have 2 km/h velocity along AO and a velocity of 2 cos 30 along the vertical direction.

22. nincompoop

I am showing you what happens if he just swam. I am not telling you that is the answer

23. anonymous

He is not swimming vertically. He is swimming along the 30 degree line.

24. Abhisar

|dw:1437537274293:dw|

25. Abhisar

It should be like this

26. anonymous

|dw:1437537285585:dw|

27. nincompoop

|dw:1437537906031:dw||dw:1437539262021:dw|

28. nincompoop

I treat them as forces and use the component method. I am not sure if 30 deg is an estimation or the actual angle.

29. nincompoop

|dw:1437540117642:dw|

30. nincompoop
31. dan815

|dw:1437542867378:dw|

32. dan815

|dw:1437542896575:dw|

33. Abhisar

Thanks all for the help.