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Homework help please. Thanks!

  1. Oct 25, 2007 #1
    I am having a very tough time trying to figure this one out. Any help would be appreciated.

    A river flows due east at 1.50 m/s. A boat crosses the river from the south shore to the north shore by maintaining a consistent velocity of 10.0 m/s due north relative to the water.

    A) What is the velocity of the boat relative to the shore?

    B) If the river is 300 m wide, how far downstream has the boat moved by the time it reaches the north shore?

    Here is a picture I made that I hope illustrates what the problem is talking about. I hope it helps you.

    [​IMG]

    One of these days I am going to have an "aha!" moment and all these problems will make sense. But that "aha!" moment is not here yet, apparently. Thanks for your help.
     
  2. jcsd
  3. Oct 25, 2007 #2

    mgb_phys

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    You have almost drawn the answer.
    At the end of the 10m/s line draw a 1.5m/s line downstream (to scale)
    Now draw a line from the start position to the end of the downstrem arrow.
    This diagonal line is the path the boat will take - it's called a force triangle.
    Then it's just a matter of simple trig.
     
  4. Oct 25, 2007 #3

    dynamicsolo

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    Look at what the current does to the boat. The boat is pointed straight north (top of page) and moves forward by 10 meters each second on the water. But the water is moving eastward (to the right) by 1.5 meters each second, carrying everything on its surface along with it. To someone watching this from the shore (or, say, hovering overhead), they see the boat traveling 10 meters to the north and 1.5 meters to the east each second. So, relative to the shore, that describes the velocity of the boat in terms of eastward (positive-x, say) and northward (positive-y) components.

    In terms of vectors for relative motion, you can express it in words this way:

    velocity of boat relative to shore =

    velocity of boat relative to water + velocity of water relative to shore .

    (In a sense, the parts involving water "cancel".)

    So now you want to ask, in the time it takes for the boat to cross the 300 meters width of the river (going north), how far will the water current have carried the boat along to the west?
     
  5. Oct 25, 2007 #4
    Using Pythagorean's Theorem is what you are suggesting? If that's the case, then the square root of 10^2 + 1.5^2 = 10.11.

    Is this the correct method?

    It might seem like common sense, but I don't want to make a silly mistake that will haunt me in the future.
     
  6. Oct 25, 2007 #5
    This still confuses me some. Would you kindly explain what I must do? I don't want to mooch, but I'm not too certain.

    Now, if this is a simple T = D/V problem and I'm just overcomplicating the subject, then boy, do I feel stupid...
     
  7. Oct 25, 2007 #6

    mgb_phys

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    In each second you go 10m across the river and you are pushed 1.5m downstream.
    How long does it take you to go across 300m at 10m/s? In that time how far will a 1.5m/s current have pushed you downstream?
     
  8. Oct 25, 2007 #7
    The answer wouldn't happen to be 45, would it?
     
  9. Oct 25, 2007 #8

    dynamicsolo

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    Show how you got it and we'll let you know... BTW, you should give answers with units if they have them.
     
  10. Oct 25, 2007 #9
    Yes, my mistake. Sorry about not including meters.

    This is what I did:

    Using the hint mgb_phys provided, I divided 300 meters by 10 meters/second. Then I multiplied the answer (30) by 1.5 meters per second, to get me my answer of 45 meters. I checked with another person who was working on the problem, and they said they got that answer too. I just want to make sure we're both not wrong, and if we are I would like to know why so I don't make the mistake again in the future.
     
  11. Oct 25, 2007 #10

    dynamicsolo

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    That's pretty much all there is to that. The downstream component of the velocity and the cross-stream (transverse) component can be dealt with separately. The hypotenuse which you found using

    10^2 + 1.5^2 = (10.11)^2 [corrected]

    gives the speed of the boat relative to the shore as 10.11 m/sec. The direction will be given from the trigonometric relation

    [tex]tan \theta = 1.5 / 10 = 0.15[/tex],

    giving a direction of about 8.5º east of due north. It wasn't clear from your original statement how they wanted the velocity relative to the shore presented. You could give it in terms of the speed and direction found here or the components I described a few posts back.
     
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