- #36
gracy
- 2,486
- 83
Ah, that's quite different.gracy said:No.And the man swims so that his velocity with respect to the water is always towards A.
Realize that since the velocity of the man with respect to the water continually changes, so that he always points towards A, his path cannot be a straight line BC.gracy said:Actually I think the following image will explain why BC is path of the swimmer/person.
Yes.Doc Al said:his path cannot be a straight line BC.
But it will be along BC but not that much straight as I have shown.Doc Al said:his path cannot be a straight line BC.
No. You forgot to include the velocity of the water with respect to the ground in your calculation. Your teacher did not forget that, which is where the u sin(theta) comes from. You need to consider the velocity of the man with respect to A, not just his velocity with respect to the water.gracy said:Because here point A is stationary,and swimmer is going towards it with velocity u along the lining the two.So his approach velocity towards A should be u - 0=u.
OK.gracy said:But as my teacher had taken his approach velocity u - u sin theta giving the following explanation.
As you can see the image of my post #30
component of river velocity in direction opposite to that of velocity of swimmer is u cos (90- theta)=u sin theta
Note:-y and x are my coordinates in the image of #30.
So over all velocity of approach (u - u sin theta) - 0=u - u sin theta
All that matters is the velocity of the man with respect to the ground. I don't know what other velocities you have in mind.gracy said:So I was having confusion whether we can include other velocities if present in the same direction or opposite to the direction of the line joining the person/swimmer and point A.
Velocity of swimmer with respect to ground=velocity of swimmer with respect to river+river velocity with (respect to ground)Doc Al said:No. You forgot to include the velocity of the water with respect to the ground in your calculation
Right.gracy said:Velocity of swimmer with respect to ground=velocity of swimmer with respect to river+river velocity with (respect to ground)
To get Velocity of approach of swimmer with respect to point A ,component of velocity of swimmer with respect to river along the line joining swimmer and point A i.e u here will have to be added with component of river velocity along the line joining swimmer and point A i.e u sintheta .
As u and u sin theta are in opposite direction there is u +(-u sin theta)- 0 =u - u sin theta.
Right?
It doesn't have to be.gracy said:But why approach velocity has to be calculated in ground frame?
That means velocity of A is only zero with respect to ground.Velocity of A with respect to swimmer/man is negative of man's velocity ,right?Doc Al said:You can also calculate it from the frame of the man, but realize that then point A is no longer stationary.
Right.gracy said:That means velocity of A is only zero with respect to ground.Velocity of A with respect to swimmer/man is negative of man's velocity ,right?
Perhaps it would help if you first looked up some examples of how Vectors are used in Force problems. Forces seem to be easier to grasp intuitively. Once you are familiar with how resolving forces can be used to solve problems easily, you could move on to velocities because the world of Vectors is the same, wherever they are applied.gracy said:But I am not getting how it gives u -u sin theta?
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?