gracy
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I did not understand.sophiecentaur said:But, if you really want to, the angle theta could be relative to the 'forward direction' of the swimmer -
I did not understand.sophiecentaur said:But, if you really want to, the angle theta could be relative to the 'forward direction' of the swimmer -
Well, what do you mean by the "frame of the swimmer"? A frame has to specify all coordinates, relative to the swimmer. The swimmer would not 'see' cartesian graph paper, lined up along and normal got the river bank. All he could do would be to put his own graph paper with x and y co ordinates parallel and at right angles to his motion through the water (or some other arbitrary orientation). I was just pointing out how difficult / unsuitable it would be to work with the swimmer's reference frame, compared with using the Earth's frame. Why not make it easy on yourself and go for the obvious way to solve this?gracy said:I did not understand.
gracy said:If I would say A is moving with velocity 5 m/s with respect to ground along the line joining AB.And B is at rest.There is some another point C which has a component of velocity 3 m/s in opposite direction to line AB. ,what is velocity approach of A with respect to B? Will it be 2 m/s?