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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?
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?

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Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 @nincompoop @dan815

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Your 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?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Yes, I know this way to solve it but I was seeking a different way to do it. Would you mind checking it?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Ok, 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
 one year ago
Best ResponseYou've already chosen the best response.0Now AB/OB = tan <AOB which gives a different answer...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0He 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.

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Yes but isn't he still making 2km/h across the vertical direction?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nope. He is travelling at 2 km/hr at 30 degrees CCW to the vertical

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So he is making progress across the river at 2cos 30 = 1.73 km/hr.

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Yes, that makes sense now.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Super. Good question.

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0what do you understand about the question

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0not the given, but the question it wants you to solve for.

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437536977448:dw

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0You 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
 one year ago
Best ResponseYou've already chosen the best response.0I am showing you what happens if he just swam. I am not telling you that is the answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0He is not swimming vertically. He is swimming along the 30 degree line.

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0It should be like this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437537285585:dw

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437537906031:dwdw:1437539262021:dw

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0I treat them as forces and use the component method. I am not sure if 30 deg is an estimation or the actual angle.

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437540117642:dw

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0here 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

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Thanks all for the help.
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