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Vector Problem

  1. Sep 25, 2008 #1
    1. The problem statement, all variables and given/known data

    a boat is a drift in a stream that flows at 3 mph 40* north of east. A wind blowing at 11 mph from a direction 50*NW also moves the boat. What will be the net direction and speed of the boat?

    2. Relevant equations

    Vwa = 3mph 40* NE
    Vwi = 11mph 50* NW

    3. The attempt at a solution

    sin50* = X/3

    Net direction will be 50* NE. Net speed will be 2.3 MPH.

    It doesnt seem right, any help?
  2. jcsd
  3. Sep 25, 2008 #2


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    Write the velocity vectors in an x-y coordinate system, say EW being x and NS being y. In xy coordinates it's easy to add them. Once you've done that, change back to the polar (magnitude-angle) representation.
  4. Sep 25, 2008 #3
    Please explain that.. I am clueless on this matter
    Last edited: Sep 25, 2008
  5. Sep 25, 2008 #4
    Take the x-components of the wind and the boat speed and add them together. Then take the y-components of each and add them together. The resulting components will be your final direction vector of the boat.

    For example, the x-component of the boats drift it's speed*cos(theta). It's y-component will be it's speed*sin(theta).

    Once you get the direction vector, take it's magnitude, and that will be the final speed of the boat.

    Edit: if you need the angle of it's final direction with respect to the NSEW coordinate system, use trig and the x,y components of the final direction vector.
  6. Sep 25, 2008 #5


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    You should also put arrows on the vectors to show what direction they are pointing. If '50*NW' means 50 degrees north of west, and if the wind is coming FROM that direction I would put the arrow on the bottom vertex. I think for the first vector the arrow should be on the upper vertex. Does that sound right? I'm trying to get the arrows to show which direction each will push the boat.

    If so take each vector and write the vertical and horizontal component of each. For the first on I get x component +3*cos(40) and y component is +3*sin(40). What do you get for the second one? Once you've done that just add x and y components to get the total vector.
  7. Sep 25, 2008 #6
    We were doing sample problems similar to this in class, and he said all we had to do was this:


    for these kind of problems.

    I will attempt to make a new grid...
  8. Sep 25, 2008 #7
    Could someone make an example grid, so I know what to follow?
  9. Sep 25, 2008 #8


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    There's methodical way to solve it by x and y components but if you want to solve it using trig on the whole triangle, then you just figure out enough angles and sides to use law of sines or law of cosines etc. Your best bet is to start with the upper angle, isn't it 110 degrees?
  10. Sep 26, 2008 #9


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    Can I do arithmetic or what? I meant 180-(50+40)=90. Not 110. This makes the problem particularly easy. Instead of those trig laws you can just use the definition of sin, cos, etc.
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