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Thread: Vector Intersection

  1. #1
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    Vector Intersection

    Hi, I have a mechanics question here I can't quite get.

    A destroyer sights a ship at a point with position vector 600(3i + j)m relative to it and moving with velocity 5j m/s. The destroyer alters course so that it moves with speed v m/s in the direction of the vector 4i + 3j. Find v so that the destroyer intercepts the ship and the time to the interception.

    Any help would be greatly appreciated.
    Thanks
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  2. #2
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    Hello steve989
    Quote Originally Posted by steve989 View Post
    Hi, I have a mechanics question here I can't quite get.

    A destroyer sights a ship at a point with position vector 600(3i + j)m relative to it and moving with velocity 5j m/s. The destroyer alters course so that it moves with speed v m/s in the direction of the vector 4i + 3j. Find v so that the destroyer intercepts the ship and the time to the interception.

    Any help would be greatly appreciated.
    Thanks
    Denote the velocity of the destroyer by $\displaystyle \vec{v_D}$ and the velocity of the ship by $\displaystyle \vec{v_S}$. Then:
    $\displaystyle \vec{v_D}=\frac{v}{5}\Big(4\vec i + 3 \vec j\Big)$, since this vector has magnitude $\displaystyle v$.
    and
    $\displaystyle \vec{v_S} = 5\vec j$
    So the velocity of the ship relative to the destroyer is
    $\displaystyle \vec{v_S}-\vec{v_D} = -\frac{4v}{5}\vec i +\Big(5-\frac{3v}{5}\Big)\vec j$
    Therefore after time $\displaystyle t$, the position of the ship relative to the destroyer is:
    $\displaystyle 600(3\vec i + \vec j) -\frac{4vt}{5}\vec i +\Big(5-\frac{3v}{5}\Big)t\vec j$
    $\displaystyle =0\vec i + 0\vec j$ when the destroyer intercepts the ship
    Solving for $\displaystyle v$ and $\displaystyle t$ gives:
    $\displaystyle t = 390$ and $\displaystyle v = \frac{75}{13}$
    But check my working!

    Grandad
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  3. #3
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    Hello, steve989!

    Got a different answer, Grandad . . . but check my work, please1


    A destroyer sights a ship at a point with position vector $\displaystyle 1800i + 600j$ relative to it
    and moving with velocity $\displaystyle 5j$ m/s.

    The destroyer alters course so that it moves with speed $\displaystyle v$ m/s
    in the direction of the vector $\displaystyle 4i+3j$

    Find $\displaystyle v$ so that the destroyer intercepts the ship and the time to the interception.
    Code:
          |
          |                     o P
          |                   o |
          |                 o   | 5t
          |               o     |
          |             o       |
          |           o         * S
          |         o           |
          |       *             |
          |  v  * |             | 600
          |   *   |3            |
          | * θ   |             |
         D* - - - + - - - - - - *
          :   4   :
          : - - - -1800 - - - - :

    The destroyer is at: $\displaystyle D(0,0)$

    The ship is at: $\displaystyle S(1800, 600)$


    The ship is moving north at 5 m/s.
    In the next $\displaystyle t$ seconds, it moves $\displaystyle 5t$ m to point $\displaystyle P.$

    . . Its position vector is: .$\displaystyle \vec S \:=\:\langle 1800,\:600+5t\rangle$


    The destroyer heads in direction $\displaystyle 4i+3j$
    Let $\displaystyle \theta$ represent its direction, where: $\displaystyle \cos\theta \,=\,\tfrac{4}{5},\;\sin\theta\,=\,\tfrac{3}{5}$

    . . Its position vector is: .$\displaystyle \vec D \;=\;\langle (v\cos\theta)t,\:(v\sin\theta)t\rangle \;=\;\left\langle\tfrac{4}{5}vt,\:\tfrac{3}{5}vt\r ight\rangle $


    The destroyer intercepts the ship at point $\displaystyle P.$
    This happens when: $\displaystyle \vec D \,=\,\vec S$

    We have: .$\displaystyle \begin{Bmatrix}\frac{4}{5}vt &=& 1800 & [1] \\ \\[-3mm] \frac{3}{5}vt &=& 600 + 5t & [2] \end{Bmatrix}$

    Divide [2] by [1]: .$\displaystyle \frac{600+5t}{1800} \:=\:\frac{\frac{3}{5}vt}{\frac{4}{5}vt} \quad\Rightarrow\quad\frac{120+t}{360} \:=\:\frac{3}{4}$

    . . $\displaystyle 480 + 4t \:=\:1080 \quad\Rightarrow\quad 4t \:=\:600 \quad\Rightarrow\quad t \:=\:150$


    Therefore, the destroyer intercepts the ship in 150 seconds.



    Substitute into [1]: .$\displaystyle \tfrac{4}{5}v(150) \:=\:1800 \quad\Rightarrow\quad v \:=\:15$

    Therefore, the destroyer must travel at 15 m/s.


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  4. #4
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    Thanks for both of your replies! I've gone though them as well as I can.

    Grandad - I tried your way and think I see how you're thinking there, I got a different formula for one stage though.

    Therefore after time , the position of the ship relative to the destroyer is:
    when the destroyer intercepts the ship
    For that bit; wouldn't it be $\displaystyle 600(3i + j) + (4/5)vti - (5 - (3/5)v)tj$ ? The signs switch as they are on different sides of the equation I assumed, but I may of miss-understood since I got a completely different answer anyway. Sorry about the layout too, I'm useless with this maths typing.

    Soroban - I do like that approach and the answers seem very reasonable, I missed the fact that the ship was moving directly north. I do have one question though for both you and grandad; why does the velocity vector of the destroyer end up with $\displaystyle v/5$ as the scalar? I thought it was just $\displaystyle v$, I think I'm going to get a simple answer to that question though.

    Thanks very much again for your answers.
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  5. #5
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    Hello everyone

    I agree with Soroban's answer: I got a sign wrong in my working. My equation for the relative velocity was correct, though. The principle is:
    The velocity of A relative to B is the velocity of A minus the velocity of B.
    So, as I said:
    $\displaystyle \vec{v_S}-\vec{v_D} = -\frac{4v}{5}\vec i +\Big(5-\frac{3v}{5}\Big)\vec j$
    not as you suggest:
    Quote Originally Posted by steve989 View Post
    ...
    For that bit; wouldn't it be $\displaystyle 600(3i + j) + (4/5)vti - (5 - (3/5)v)tj$ ? The signs switch as they are on different sides of the equation I assumed, but I may of miss-understood since I got a completely different answer anyway. Sorry about the layout too, I'm useless with this maths typing.

    However, I then got a sign wrong when I equated the $\displaystyle \vec j$ component of the displacement to zero. The equations should have been:
    $\displaystyle \vec i$ component: $\displaystyle 1800 - \frac{4vt}{5} = 0$

    $\displaystyle \vec j$ component: $\displaystyle 600 + 5t -\frac{3vt}{5}=0$
    which, of course, are the same as Soroban's equations [1] and [2].
    Quote Originally Posted by steve989 View Post
    ...I do have one question though for both you and grandad; why does the velocity vector of the destroyer end up with $\displaystyle v/5$ as the scalar? I thought it was just $\displaystyle v$, I think I'm going to get a simple answer to that question though.

    Thanks very much again for your answers.
    Because the magnitude of $\displaystyle 4\vec i + 3\vec j$ is $\displaystyle 5$. So $\displaystyle v(4\vec i + 3\vec j)$ will have magnitude $\displaystyle 5v$.

    Grandad
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  6. #6
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    Ahh I see, all makes perfect sense now, I knew there had to be simple reasoning behind that. Thanks a lot to you both.
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