How Is Drift Calculated in River Boat Problems?

[Doc Al]

A man and river are moving with same velocity u..

I assume you mean that the man moves with speed u with respect to the river, and the river moves with speed u with respect to the ground.

A person is at point B initially.His velocity with respect to river is always towards point A.River has velocity in horizontal direction.So that person's path is shown as BC.At one instant D I have shown person's velocity withg respect to river as u and component of river's velocity as u Sin theta.
{cos (90-theta) =sin theta.}

I assume the river moves to the right, call it the +x direction. And the man swims so that his velocity with respect to the water is always up, call that the +y direction. So his velocity with respect to the ground is $u\hat{x} + u\hat{y}$. And since he's moving away from B, that will be his separation velocity from B. (He's moving away from B, not toward B.)

If I have the wrong setup, let me know.

So my teacher said that velocity of approach of person towards B is u-u sin theta.

I do not see where that came from.

Shouldn't then velocity of approach of person towards B be "u"?

Not if I understand the problem statement.

I'm not sure I understand all the arrows in your diagram.

 

[gracy]

I assume you mean that the man moves with speed u with respect to the river, and the river moves with speed u with respect to the ground.

Yes ,your assumption is correct.

 

[gracy]

And the man swims so that his velocity with respect to the water is always up,

No.And the man swims so that his velocity with respect to the water is always towards A.

 

[gracy]

So that person's path is shown as BC.

Actually I think the following image will explain why BC is path of the swimmer/person.
I have changed this a little bit.That's why refer image of my post #36 rather than the one below.

 

[gracy]

So my teacher said that velocity of approach of person towards B is u-u sin theta.

I do not see where that came from.

That's what I wanted to ask ,so I asked in my post #13

So 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 ,what is velocity approach ,of A with respect to B? it will be 5 m/s ?

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.
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
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.
That's why I asked the below question in my post # 19

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?

 

[gracy]

 

[Doc Al]

No.And the man swims so that his velocity with respect to the water is always towards A.

Ah, that's quite different.

Actually I think the following image will explain why BC is path of the swimmer/person.

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]

his path cannot be a straight line BC.

Yes.

 

[gracy]

Did you get what am I actually asking?

 

[gracy]

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]

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.

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.

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

OK.

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.

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]

No. You forgot to include the velocity of the water with respect to the ground in your calculation

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?

 

[gracy]

But why approach velocity has to be calculated in ground frame?

 

[Doc Al]

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?

Right.

But why approach velocity has to be calculated in ground frame?

It doesn't have to be.

You can also calculate it from the frame of the man, but realize that then point A is no longer stationary.

 

[gracy]

You can also calculate it from the frame of the man, but realize that then point A is no longer stationary.

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]

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?

Right.

 

[gracy]

But I am not getting how it gives u -u sin theta?

 

[sophiecentaur]

But I am not getting how it gives u -u sin theta?

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]

I am really not getting from where u-usin theta is coming in frame of the swimmer.

 

[sophiecentaur]

I would not try to solve such a problem in the frame of the swimmer, in any case. But, if you really want to, the angle theta could be relative to the 'forward direction' of the swimmer - which would be the only logical reference axis if you are truly working in the swimmer's frame. 'The system' will allow you to choose any frame you like but why not choose the most convenient one (i.e. the Earth)?

 

[gracy]

But, if you really want to, the angle theta could be relative to the 'forward direction' of the swimmer -

I did not understand.

 

[sophiecentaur]

I did not understand.

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?

 

[Sashwat Tanay]

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?

velocity approach of A with respect to B is 5 m/s. I don't think point C will matter.