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Vectors - Boat & Stream problem

  1. Aug 8, 2012 #1
    The problem statement, all variables and given/known data

    A boatman wants to cross a canal that is 3 km wide and wants to land at a point 2 km upstream from his starting point. The current in the canal flows at 3.5 km/h and the speed of his boat is 13 km/h.
    (a) In what direction should he steer?
    (b) How long will the trip take?

    This is not for an actual course...this is from a textbook, but I am self teaching. I am having an issue finding out how they solved for this. I need a solution to this so I can understand it in the future.

    The answers:

    (a) At an angle of 43.4 deg from the bank, toward upstream
    (b) 20.2min
     
  2. jcsd
  3. Aug 8, 2012 #2
    Your destination is across and upstream, 3km north(y=3) and 2km east(x=2).
    Your resultant velocity in direction tan-1(3/2).
    This resultant velocity is made up of your velocity plus river current velocity.
    You can resolve your boatman velocity into vertical(y) and horizontal(x) to obtain the magnitude and direction required.
     
  4. Aug 8, 2012 #3
    There may be a quicker way to do this, and I hope there is since you posted it in the calculus and beyond section, but off the top of my head the obvious one is to simply draw a right triangle with sides (3) and (3.5t+2) with hypotenuse (13t). Just use your standard pythagorean theorem to get a quadratic...

    I got something like 156.75t^{2}-14t-13 = 0

    Use the quadratic formula, which is a little messy here, but no big deal if you are allowed to use even a basic calculator on the problem. Solving this for the positive t value gives you something like t =.336 hrs = 20.165 min

    Just plug in this t value to get the values of the sides of the original triangle, and now you can use any of your basic trig functions to find the angle you need.

    To explain the set up: hopefully you see why I set up the triangle this way, but if you don't, it's very simple. Rate x Time = Distance, so we can use the given current and boat speed rates, multiply them by the unknown t to get distances. 13t is the distance traveled by the boat relative to the surface of the water. 3.5t is the distance the surface of the water "shifts" the boat downstream during the trip. And then of course 2 and 3 are distances between/on the stationary shores.
     
  5. Aug 8, 2012 #4
    Thank-you bossman, and you solved it the way they probably wanted (or in a similar fashion). This is from a chapter about vectors etc. This shall help me whenever I come across anything similar to this...I kinda feel stupid that I didn't figure it out though.
     
  6. Aug 8, 2012 #5
    No worries, I'm learning much more advanced math now and it still took me a minute to see which way to set it up. The takeaway here is just to remember that you can often set up problems easily by interconverting rates, times and distances, using the unknown one as a variable. In this case we can just convert the rates into distances with a variable for time, etc. etc.

    That said, do be mindful of where you're posting these questions. This question belongs in the precalculus or introductory physics section.
     
  7. Aug 8, 2012 #6
    Like to know is there reasons for sequence of the questions.
    Do the first is easier to solve than the next.
    I'm doing self teaching too.
     
  8. Aug 9, 2012 #7
    Although the questions in textbooks are usually ordered with intent, as may be the case here since there is a way you can find the angle first, it's good practice to look at problems like this as being one situation with two (or more) unknowns and then figuring out for yourself the best, or most efficient way to find solutions. If a textbook constantly asks you to use inefficient methods, it's probably not a very good book.
     
  9. Aug 10, 2012 #8
    This is from Stewart 7E Calculus textbook in Chapter 12 section 2 problems. It is over vectors; this is the first problem I came across in the practice questions I had issues with, mainly because I'm a bit rusty on my solution from a Physics standpoint.
     
  10. Aug 10, 2012 #9
    Boatman upstream component velocity = vbCosθ
    Boatman across stream component velocity = vbSinθ

    River downstream velocity=-vr

    (vbCosθ-vr)t=2 ........(1)

    (vbSinθ)t=3 .........(2)
     
    Last edited: Aug 10, 2012
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