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Thread: relative velocity interception

  1. #1
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    relative velocity interception

    a sailing boat is travelling at a constant speed of 4 km hr^-1 in a direction 35 degrees north of east. a motorboat wishes to intercept the sailing boat is travelling at a speed of 10 km hr^-1. at a certain point the motorboat is 2km northwest of the sailing boat. in what direction should the motor boat head in order to intercept the sailing boat

    what ive got so far: velocity of sailing boat=3.27i+2.3j velocity of motor boat=(10cosx)i+(10sinx)j if i get there relative velocity the j components should cancel out so 2.3=10sinx x=13.297 degrees sadly this is not the right answer. the right answer is 21.8 degrees south of east.
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    Re: relative velocity interception

    Let the motorboat be at the origin at $t=0$ with the saiboat 2km southeast of the motorboat at position $r_s=2\cos(-45)i + 2\sin(45)j$.

    Sailboat velocity ... $v_s=4\cos(35)i + 4\sin(35)j$.

    Motorboat velocity ... $v_m=10\cos{\theta} i + 10\sin{\theta} j$

    Let $t$ be the time of interception.

    In the E/W direction ...

    $10\cos{\theta} \cdot t = 2\cos(-45) + 4\cos(35) \cdot t \implies t = \dfrac{\cos(-45)}{5\cos{\theta}-2\cos(35)}$

    In the N/S direction ...

    $10\sin{\theta} \cdot t = 2\sin(-45) + 4\sin(35) \cdot t \implies t = \dfrac{\sin(-45)}{5\sin{\theta}-2\sin(35)}$

    Since $\sin(-45) = -\cos(-45)$ ...

    $5\cos{\theta} - 2\cos(35) = 2\sin(35) - 5\sin{\theta}$

    $\cos{\theta}+\sin{\theta} = \dfrac{2}{5}\bigg[\cos(35)+\sin(35) \bigg]$

    Let the right side = constant, $k$, and square both sides ...

    $\sin(2\theta) + 1 = k^2$

    $\theta = \dfrac{1}{2} \arcsin(k^2-1) \approx -21.8^\circ$ or $21.8^\circ$ S of E.
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    Re: relative velocity interception

    im also trying to solve in a different way cause sometimes you are not given the distance between the ships. here is a example of the way im trying to do it from a different question. for the way they are working it out could you use the same way for the question i posted.?
    relative velocity interception-win_20170212_16_45_36_pro.jpg relative velocity interception-win_20170212_16_46_36_pro.jpg im trying to work it out this way but it wont work
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    Re: relative velocity interception

    the text re-orients the i and j directions as shown in the diagram ...
    Attached Thumbnails Attached Thumbnails relative velocity interception-vector1.png  
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    Re: relative velocity interception

    Thanks that's what I was looking for. I kept getting my answer 23.2 why are you taking 45 degree away? Is it to get it back to normal I and j components
    Last edited by markosheehan; Feb 12th 2017 at 09:00 AM.
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    Re: relative velocity interception

    Quote Originally Posted by markosheehan View Post
    Thanks that's what I was looking for. I kept getting my answer 23.2 why are you taking 45 degree away? Is it to get it back to normal I and j components
    Reorientation of the i direction from its normal direction East to a new direction Southeast is equivalent to a $-45^\circ$ rotation ... so, the new direction for $\vec{i}$ is $-45^\circ$. Since $\theta$ is relative to that new direction, the direction of $\theta$ relative to due East is actually $-45^\circ + \theta$.

    Since $\theta > 0$, I just subtracted $45^\circ$
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