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Your sketch is accurate. You are after
Yes, I know this way to solve it but I was seeking a different way to do it. Would you mind checking it?
Sure. Go ahead
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|
Now AB/OB = tan
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
Yes but isn't he still making 2km/h across the vertical direction?
Nope. He is travelling at 2 km/hr at 30 degrees CCW to the vertical
So he is making progress across the river at 2cos 30 = 1.73 km/hr.
Yes, that makes sense now.
Super. Good question.
what do you understand about the question
not the given, but the question it wants you to solve for.
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
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.
I am showing you what happens if he just swam. I am not telling you that is the answer
He is not swimming vertically. He is swimming along the 30 degree line.
It should be like this
I treat them as forces and use the component method. I am not sure if 30 deg is an estimation or the actual angle.
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
Thanks all for the help.