How Is Drift Calculated in River Boat Problems?

[gracy]

What is drift in river boat problems?Is it velocity of river divided by time taken o reach the point aimed?Or velocity of river-countering velocity of swimmer or boat in the direction of boat velocity divided by time taken o reach the point aimed?In the below video ,

although the accent is really hard to understand I found content quite good.So I was watching this but I encountered this doubt.
While calculating drift at time 17:45 he took velocity as rivers velocity-it's countering velocity multiplied by time
but in other problems such as at time 6:22 in calculation of drift only velocity of river (no countering velocity)multiplied by time is taken .Why?
I think it is just a mistake.That's why I will not be able to show any effort.But if you find that it's correct,just give me a hint so that I can clear my doubt on my own.

 

[NascentOxygen]

Hi gracy. I won't dare click on a 40 min video, it would consume a sizeable chunk of my tablet's monthly data plan!

I don't like the vector diagram shown as the frontspiece to that video. It flouts the usual convention for labelling in these questions, so in this its author is, IMHO, unforgiveable! It is just going to confuse you, and me, too.

Normally we need memorise and write just one general equation, viz., v_b = v_{b rel w} + v_w

However, your video has labelled v_{b rel w} as v_b so I think you should immediately relabel that to v_{b rel w} and then use the equation I just gave.

As for drift? Well, if the boat's engine were to be switched off so the boat has no velocity relative to the water, then the boat's motion (relative to the land) after that would be its drift. So drift is motion caused by the water's velocity, so it has a speed and a direction identical with that of the water.

This means the total drift will be river velocity times time.

That's how I see it, and I hope your video treats it the same way.

 

[anorlunda]

I don't think that term drift is very well defined in boating problems. I would have to watch the video to understand what the author meant. It's too long. Sorry.

I recommend that you make your own diagrams and your own nomenclature, and figure it out for yourself.

 

[NascentOxygen]

Yes, anorlunda you do make a good point. "Drift" is used in two senses. I described one where the boat's engine is used to counter the drift.

The other sense is that drift can be the distance the boat is moved by the river flow even when the boat is using its engine.

I can use the word in both its senses: the experienced ferry captain used his engines to counter the water's drift so that he crossed directly to the opposite wharf with barely a metre drift downstream.

 

[anorlunda]

I can use the word in both its senses: the experienced ferry captain used his engines to counter the water's drift so that he crossed directly to the opposite wharf with barely a metre drift downstream.

Well said. I call it swim like a duck or swim like a dog.

 

[sophiecentaur]

I think it would be very unusual to find that the terms used in a practical subject like marine navigation correspond rigidly to terms used in Physics. If that video seems to depart from what you would expect, talking in strict Physics terms then I think you just need to accept it. As long as you understand the Physics of what's going on and know, in this case, the velocity vector of the river current and the velocity vector os the boat then you can work out the resultant velocity vector. An expression like "counter the ferry drift" is just a description for what a skipper does without formally using vectors. Life's too short to get involved with how the vernacular ties in with formal Scientific theories.

 

[gracy]

Drift in river boat problems is
Say if there is a boat which has velocity 5m/s with respect to the river. It aims to reach point A, however, because of the river speed it reaches another point B.Then this Distance AB is called 'drift.So it's a distance.

 

[gracy]

Relative velocity of approach /SEPARATION
Let's say if there is a point A and point B.My teacher says relative velocity of approach is always along the line joining the A and B.

As 5 cos 30>2 cos 60
it is case of approach
relative velocity of approach would be
4.3-1=3.3 m/s
but if B velocity along the line joining A and B would have been less than that of A,It would have been considered as case of separation.
My question what if B is stationary.My teacher still took their relative velocity of approach
I did not understand why?As B is stationary there should not be any relative velocity.

 

[sophiecentaur]

relative velocity of approach is always along the line joining the A and B

When you are dealing with this sort of problem you need to realize that there are two different sets of vectors at work. There are displacement vectors - which for a boat is its Course Over the Ground (COG) and velocity vectors, of which there are two: Flow vector of the water and velocity of the boat through the water (Speed and Heading). It is easy to confuse the two vectors and, if you draw them on the same diagram, you have to avoid drawing wrong conclusions about 'where' the velocity vector is - it isn't anywhere. In a simple case like this, there is no acceleration so the displacement will lie on a straight line as time progresses.

As B is stationary there should not be any relative velocity

If one is moving relative to the other so there has to be a finite relative velocity. The only case with no relative velocity would be if the boat were heading directly upstream with a velocity relative to the water that is equal to the velocity of the water over the ground (for instance, when the boat is approaching a mooring buoy and comes to a halt just before hooking on).

 

[gracy]

I don't understand.I just wanted to ask A is moving with velocity 5 m/s with respect to ground.And B is at rest ,what is relative velocity of A with respect to B?I think it will be 5 m/s .

 

[sophiecentaur]

I don't understand.I just wanted to ask A is moving with velocity 5 m/s with respect to ground.And B is at rest ,what is relative velocity of A with respect to B?I think it will be 5 m/s .

It will depend upon the position at the time. If the velocity is along the line of AB then you are right but it could be anything, for different positions and directions.

 

[Doc Al]

I don't understand.I just wanted to ask A is moving with velocity 5 m/s with respect to ground.And B is at rest ,what is relative velocity of A with respect to B?I think it will be 5 m/s .

Sure, in that case the relative velocity of A with respect to B would be 5 m/s. But that's not the same as the velocity of approach, which depends on the direction of that relative velocity.

 

[gracy]

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 ?

 

[Doc Al]

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 ?

Yes, exactly. And if A were moving perpendicular to the line AB, then the approach velocity (at that instant) would be zero.

 

[gracy]

Thanks.

 

[gracy]

Can you please answer my post #1.

 

[gracy]

I have explained in my post # 7 what "drift "is in my problem's context.

 

[gracy]

If two bodies have same velocity i.e same magnitude as well as direction then relative velocity of anyone body with respect to the other will be zero,right?

 

[gracy]

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?

 

[NascentOxygen]

If two bodies have same velocity i.e same magnitude as well as direction then relative velocity of anyone body with respect to the other will be zero,right?

Yes.

Taking the general equation
v_a + v_{b rel a} = v_b

Re-arranging,
v_{b rel a} = v_b - v_a

= 0 because the difference between identical vectors is zero.

 

[Doc Al]

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.There is some another point C which has a component of velocity 3 m/s in opposite direction to line AB.

So A, B, and C are in a line. B is at rest with respect to the ground. A moves toward B with a speed of 5 m/s with respect to the ground. C moves away from B with a speed of 3 m/s with respect to the ground. Correct?

what is velocity approach of A with respect to B? Will it be 2 m/s?

I think you mean the approach velocity of A with respect to C, right? Which would be 2 m/s.

 

[gracy]

I think you mean the approach velocity of A with respect to C, right

Right.

 

[gracy]

NO.Sorry I didn't understand at first glance.I meant approach velocity of A with respect to B.I thought you are talking about my mistake Velocity approach in place of approach velocity.

 

[Doc Al]

NO.Sorry I didn't understand at first glance.I meant approach velocity of A with respect to B.

Then why do you mention C and its velocity at all? C is irrelevant to the relative velocity of A and B.

 

[gracy]

C is irrelevant to the relative velocity of A and B.

And what about approach velocity?Is C also irrelevant to the approach velocity of A and B?

 

[Doc Al]

And what about approach velocity?

C has nothing to do with the approach velocity of A and B. Why would you think it does?

 

[gracy]

Why would you think it does?

Because my instructor said so.

 

[Doc Al]

Because my instructor said so.

That's a bit hard to believe.

Could it be that you misremember what he said?

For example: You drive your car at 60 mph toward a tree. What's your approach velocity with respect to the tree? Can you answer that? Of course, you can. Notice that I didn't mention anyone or anything else. The fact that something else might be moving in some direction at some speed has no bearing on how fast you are approaching that tree.

 

[gracy]

That's a bit hard to believe.

Well.I can explain you what my instructor said.You can point out if you find any mistake in my interpretation..
It was river swimmer problem.
A man and river are moving with same velocity u..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.}
So my teacher said that velocity of approach of person towards B is u-u sin theta.

C has nothing to do with the approach velocity of A and B

Shouldn't then velocity of approach of person towards B be "u"?
That's why I mentioned about c in my question.

 

[gracy]

This is correct picture of my river swimmer problem.

 

[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)?