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Math Help - Applications of Vector Addition

  1. #1
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    Applications of Vector Addition

    Hi, I need help with another vector word problem.

    A supply boat needs to cross a river from point A to point B, as shown in the attached diagram. Point B is 1.5 km downstream. The boat can travel at a speed of 20 km/h relative to the water. The current is flowing at 12 km/h. The width of the river is 500 m.

    1) Determine the heading the captain should set to cross the river to point B.
    2) Determine the heading the captain must set to return to point A.

    I'm not really sure what I'm doing wrong with this one. I've tried it a couple of times, and keep getting different wrong answers. I don't know why, but word problems always seem to mess me up. So if anyone can offer some advice or a solution, it would be appreciated.

    Thanks
    Attached Thumbnails Attached Thumbnails Applications of Vector Addition-20100214233322.png  
    Last edited by sam314159265; February 14th 2010 at 08:05 PM. Reason: wrong file
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  2. #2
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    I'm not sure what "1.5km downstream" means. Is it 1.5KM relative to Point A, or what?

    As for the actual problem: The boat is moving from point A to B, at a speed of 20km/h. We can represent its vector as AB=20. The vector parallel to the river bank we can say is I dunno RC (river current), and is equal to 12. If the captain takes his boat from point A to B, he will be pushed off course away from B, by the current, since if you add AB to RC, you get a vector ABRC that is equal to the actual direction and magnitude of the boat. The captain should then direct the boat to the right of Point B, in order to get a total vector that points from A to B.
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  3. #3
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    Quote Originally Posted by sam314159265 View Post
    Hi, I need help with another vector word problem.

    A supply boat needs to cross a river from point A to point B, as shown in the attached diagram. Point B is 1.5 km downstream. The boat can travel at a speed of 20 km/h relative to the water. The current is flowing at 12 km/h. The width of the river is 500 m.

    1) Determine the heading the captain should set to cross the river to point B.
    2) Determine the heading the captain must set to return to point A.

    I'm not really sure what I'm doing wrong with this one. I've tried it a couple of times, and keep getting different wrong answers. I don't know why, but word problems always seem to mess me up. So if anyone can offer some advice or a solution, it would be appreciated.

    Thanks
    1. Define a coordinate system: The bank of A as x-axis, perpendicular to the bank through A the y-axis.

    2. The direction AB is the vector \overrightarrow{AB} = (-1.5, 0.5)
    The vector describing the current is \vec c = (-12, 0)

    3. You are looking for a vector \vec h = (x,y) such that

    \vec c + \vec h = \overrightarrow{AB}

    Solve for \vec h
    Attached Thumbnails Attached Thumbnails Applications of Vector Addition-boot_ueberfluss.png  
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  4. #4
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    Set up coordinate system so that the positive x-axis is up river and the positive y axis is at right angles to the river with the origin at the point where the boat starts, point A. If the boat is headed at angle \theta to the line directly across the river, then its velocity vector, relative to the water, is given by <20cos(\theta), 20sin(\theta)>. The velocity vector of the river is <-12, 0>. The actual velocity is [tex]<20 cos(\theta)- 12, 20sin(\theta)> so that after t hours, the boat will have gone <20 cos(\theta) t- 12t, 20sin(\theta)t>. To get to point B when [tex]<(20 cos(\theta)- 12)t, 20sin(\theta)t>= <-1.5, .5> so you have two equations, 20 cos(\theta)- 12)t= -1.5 and 20sin(\theta)t= .5 to solve for [itex]\theta[/itex] and t. Since you are not asked what t is, I recommend dividing one equation by the other to eliminate t and have a single equation for \theta.
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  5. #5
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    How would I divide one equation by the other, in order to eliminate t? and why was t included? Sorry about all the questions, but I am still confused...
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  6. #6
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    Quote Originally Posted by sam314159265 View Post
    How would I divide one equation by the other, in order to eliminate t? and why was t included? Sorry about all the questions, but I am still confused...
    \begin{array}{rcl}20 (cos(\theta)- 12)t&=& -1.5 \\<br />
20sin(\theta)t &=&  .5\end{array}

    Divide both left hand sides and both right hand sides:

    \dfrac{\cos(\theta)-12}{\sin(\theta)} = -3

    Now solve for \theta
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