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Drift in river boat problems

  1. Apr 8, 2015 #1
    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.
     
    Last edited: Apr 8, 2015
  2. jcsd
  3. Apr 8, 2015 #2

    NascentOxygen

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    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., vb = vb rel w + vw

    However, your video has labelled vb rel w as vb so I think you should immediately relabel that to vb 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.
     
  4. Apr 8, 2015 #3

    anorlunda

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    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.
     
  5. Apr 8, 2015 #4

    NascentOxygen

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    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.
     
  6. Apr 8, 2015 #5

    anorlunda

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    Well said. I call it swim like a duck or swim like a dog.
     
  7. Apr 8, 2015 #6

    sophiecentaur

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    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.
     
  8. Apr 8, 2015 #7
    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.
    exampleriverlast.png
     
  9. Apr 9, 2015 #8
    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.
    upload_2015-4-9_11-12-37.png

    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.
     
  10. Apr 9, 2015 #9

    sophiecentaur

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    When you are dealing with this sort of problem you need to realise 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.
    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).
     
  11. Apr 9, 2015 #10
    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 .
     
  12. Apr 9, 2015 #11

    sophiecentaur

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    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.
     
  13. Apr 9, 2015 #12

    Doc Al

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    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.
     
  14. Apr 9, 2015 #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 ?
     
  15. Apr 10, 2015 #14

    Doc Al

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    Yes, exactly. And if A were moving perpendicular to the line AB, then the approach velocity (at that instant) would be zero.
     
  16. Apr 10, 2015 #15
    Thanks.
     
  17. Apr 10, 2015 #16
    Can you please answer my post #1.
     
  18. Apr 10, 2015 #17
    I have explained in my post # 7 what "drift "is in my problem's context.
     
  19. Apr 10, 2015 #18
    If two bodies have same velocity i.e same magnitude as well as direction then relative velocity of any one body with respect to the other will be zero,right?
     
  20. Apr 10, 2015 #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?
     
    Last edited: Apr 10, 2015
  21. Apr 10, 2015 #20

    NascentOxygen

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    Yes.

    Taking the general equation
    va + vb rel a = vb

    Re-arranging,
    vb rel a = vb - va

    = 0 because the difference between identical vectors is zero.
     
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