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Math Help - Intercepting vectors

  1. #1
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    Intercepting vectors

    Rower A is stationary at the point A adjacent to the river bank when he spots an object floating down the river. He estimates that the object is 50m away and heading south at a speed of 3m/s (the same speed as the current) see diagram below.

    Assuming rower A can sustain a rowing speed of 4m/s

    1 Draw a diagram of the velocity vectors involved if rower a is to intercept the object

    2 calcuate the direction with respect to the river bank he needs to row (in degrees) in order to achieve his goal.

    Now part 1 of the question i know how to do, but with part 2 i'm stuck, not really sure where to begin. If someone could just give a few hints then hopefully I can take it from there.

    Many thanks
    Attached Thumbnails Attached Thumbnails Intercepting vectors-diagram.jpg  
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  2. #2
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    let \theta be the desired angle for rower A's velocity w/r to the river bank.

    using the diagram as reference, the rower's desired horizontal displacement is 50\sin(20) meters.

    rower's velocity in the horizontal direction would be ...
    v_x = 4\sin{\theta}

    so ...

    4\sin{\theta} \cdot t = 50\sin(20)

    t = \frac{50\sin(20)}{4\sin{\theta}}


    in the vertical direction relative to the shore, the object's distance traveled + the rower's distance traveled must sum to be 50\cos(20) meters ...

    relative to the shore in the vertical direction ...
    object's speed = 3 m/s
    rower's speed = 4\cos{\theta} - 3 m/s

    3t + (4\cos{\theta}-3)t = 50\cos(20)

    4\cos{\theta} \cdot t = 50\cos(20)

    all that is left is to sub in the first expression for t into the above equation and solve for \theta ... which, btw, is really easy in this particular case.
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  3. #3
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    Many thanks, I'll not pretend to have totally grapsed everything you've said, but i am working my way through bit by bit. Its coming abiet a bit slowly!

    thanks again
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  4. #4
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    Hello, ally79!

    Rower A is stationary at the point A adjacent to the river bank
    when he spots an object floating down the river.
    He estimates that the object is 50 m away and heading south at a speed of 3 m/s.

    Assuming rower A can sustain a rowing speed of 4 m/s,

    1) Draw a diagram of the velocity vectors involved if rower A is to intercept the object.

    2) Calculate the direction with respect to the river bank he needs to row
    (in degrees) in order to achieve his goal.
    Code:
          N           * B
          |          *|
          |         * |
          |        *  |
          |    50 *20| 3t
          |      *    |
          |     *     |
          |    *      |
          |20*       * C
          |  *     *
          | *θ  * 4t
          |* *
        A *
          |

    We are given: . \angle NAB = 20^o,\;AB = 50
    . . Note that: \angle ABC = 20^o

    In t seconds, A has moved 4t m to point C.

    In t seconds, B has moved 3t m to point C.

    Let \theta = \angle BAC

    We want \angle NAC.


    Using the Law of Sines: . \frac{\sin\theta}{3t} \:=\:\frac{\sin20^o}{4t} \quad\Rightarrow\quad \sin\theta \:=\:\frac{3\sin20^o}{4} \:=\:0.256515107

    Hence: . \theta \:=\:14.86338095^o \:\approx\:14.86^o


    Therefore: . \angle NAC \;=\;14.86^o + 20^o \;=\;\boxed{34.86^o}



    Edit: skeeter pointed out an error . . . *blush*
    I forgot: the boat is also affected by the current.

    I'll have to work on a corrected version.

    Sorry . . .
    .
    Last edited by Soroban; September 19th 2008 at 05:49 PM.
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  5. #5
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    Skeeter,
    I follwed through what you said and I was able to follow it and understand it, Hower when substuting in t and rearranging to solve for theta I ended up with 50 Sin (20)/50cos (20) = 4 Sin theta/4 Cos Theta

    Which i cancelled down to Tan 20 = Tan Theta Which gives me the original 20 degree angle.

    If anyone can point out where I went wrong i'd be grateful!
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  6. #6
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    ironic, isn't it? but correct.
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  7. #7
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    is that a further 20 degrees on top of the 20 degrees already?
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  8. #8
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    How can the angle of interception be 20 degrees when the initial angle is 20 degrees? Surely as the object is floating down stream the angle will be increasing every second?

    Can someone please explain this to me? I apologise if i'm missing something obvious
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  9. #9
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    Quote Originally Posted by ally79 View Post
    is that a further 20 degrees on top of the 20 degrees already?
    no ... the 20 degrees relative to the shore is what the boat must steer to intercept the object.
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