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?

- Abhisar

- chestercat

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- Abhisar

- Abhisar

|dw:1437535101502:dw|

- anonymous

Your sketch is accurate. You are after

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## More answers

- Abhisar

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

- anonymous

Sure. Go ahead

- 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|

- Abhisar

Now AB/OB = tan

- Abhisar

- 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

- Abhisar

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

- anonymous

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

- anonymous

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

- Abhisar

Yes, that makes sense now.

- anonymous

Super. Good question.

- Abhisar

Thank you.

- nincompoop

what do you understand about the question

- nincompoop

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

- anonymous

You're welcome

- 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

- nincompoop

|dw:1437536977448:dw|

- 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.

- nincompoop

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

- anonymous

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

- Abhisar

|dw:1437537274293:dw|

- Abhisar

It should be like this

- anonymous

|dw:1437537285585:dw|

- nincompoop

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

- 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.

- nincompoop

|dw:1437540117642:dw|

- nincompoop

here are extra readings to do
http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Vectors/VectorAdditionSteps.html
http://physics.bu.edu/%7Eduffy/semester1/c3_vadd_comp.html

- dan815

|dw:1437542867378:dw|

- dan815

|dw:1437542896575:dw|

- Abhisar

Thanks all for the help.

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