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Two vector problems

  1. Sep 30, 2009 #1
    1) Dan is crossing a river from point A to point B. The distance between the points are 60 meters, and B are straight on the other side of the river from A. The river current is 0.5 meters / second towards the right and Dan's rowing speed is 2 m/s.

    Find the point C Dan has to aim for to end up at point B.


    I've found the following:
    - He will use 30s ( 60m / 2ms ) rowing across the river.
    - If he crosses the river directly towards B he will end up 15m (30s * 0.5ms) to the right of B.
    - The speed of the boat when rowing directly towards B is ~= 2,06ms ((2^2 + .5^2)^.5)

    But I cant seem to formulate a good recipe for finding the point C.

    2) Select two points A and B on the X-axis. Find the point P which divides the line segment AB into 1:2, that is so AP : PB = 1 : 2.

    Using vectors, show that the X coordinate of the point P can be described by (2A + B) / 3.


    I can see that the relationship can be described as "OP = OA + 1/3 AB" or "OP = OB - 2/3AB" but nothing that i can manipulate until i get it into the form (2A + B) / 3.
     
  2. jcsd
  3. Sep 30, 2009 #2

    CompuChip

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    First (in any vector problem): draw a picture!

    Hint: look at the velocity vectors. Set up a coordinate system and draw the vectors vDan (of which you know the magnitude) and vriver (of which you know the magnitude and direction). Since you also know the magnitude and direction of the "real" velocity vector v = vDan + vriver, you should be able to solve for the direction of vDan.
     
  4. Sep 30, 2009 #3

    CompuChip

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    For the second question, your start looks right.
    Now introduce three vectors P (from the origin to the point P), A (from the origin to the point A) and B (from the origin to the point B) and express your last equality (OP = OA + (1/3) AB) in terms of these vectors.
     
  5. Sep 30, 2009 #4
    CompuChip

    Thanks. I've solved the second problem now, using OA and OB as you suggested instead of OA and AB did the trick.

    I am still stuck on the first problem however. I do have a drawing (in fact, I have several :p) but I don't have a scanner. In ascii it looks something like this:

    Code (Text):

    -------------C------B--------------------
    ^                      R
    |                    ----->
    |                   ^   /\
    60m                 |   /
    |                  D|  / X
    |                   | /
    v                   |/
    --------------------A--------------------
     
    A and B are the starting and (attempted) landing position,
    D = speed and direction of Dan (before accounting for current)
    R = speed and direction of river
    X = speed and direction of Dan (when accounting for the current)

    I know X, but I don't see how I can translate this into the direction I need towards C. I want the direction I take towards C to effectively cancel the current, but I cant seem to bend my head around how to do it :/

    k
     
  6. Sep 30, 2009 #5
    try to draw D towards C and X towards D. Suppose D and X make an angle phi.

    choose phi so that the upriver component of D is canceled by R.
     
  7. Oct 1, 2009 #6
    I don't get it :/

    X towards D? From where? And the velocity of X is determined by both D and R, so if I move D, then X is invalidated.

    k
     
  8. Oct 1, 2009 #7
    this is what i think you do...

    draw a triangle with 2m/s as the hyp and 0.5m/s as the adjacent. The opisite is the speed he will cross the river ..

    now calculate the angle of the triangle. it is about 75 degrees.. .. now u no the angle that he needs to cross at.....

    draw another triangle with the oppisite side 60 m... and solve for the adjacent... it works out about 15.49 meters.. so he has to aim 15.49 m up river of the intended target
     
    Last edited: Oct 1, 2009
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