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  1. #1
    Member classicstrings's Avatar
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    Question Difficulty

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  2. #2
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    earboth's Avatar
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    nr. 6 and nr. 7 missing

    Hi,

    to 1) The fishing boat needs \frac{400}{8.5}\approx 47\ s
    the ferry needs \frac{120}{3}= 40\ s

    So the ferry is 21 m south of the fishing boat when it reaches the point X. (Quite a narrow escape!)

    to 2) Use the Pythagoran rule: d=\sqrt{400^2+120^2}\approx  417.6 m

    to 3) Use the Pythagoran rule. One leg is 400-8.5*t and the other is 120-3*t. So you get:
    d(t)=\sqrt{(400-8.5t)^2+(120-3t)^2}

    to 4) You are looking for the minimum distance. So d(t) has a minimum if D(t)=(d(t)) has a maximum. For convenience I use D(t)
    D(t)=81.25t^2-7520t+174400
    \frac{d(D(t))}{dt}=162.5t-7520. Thus t ≈ 46.3 s.

    That means 0.7 s before the fishing boat arrives at point X the vessels are closest to each other.

    d(46.3) ≈ 19.97 m ≈ 20 m

    to 5) The visibility is 50 m. So the distance is less or equal 50 m.

    d(t)=\sqrt{(400-8.5t)^2+(120-3t)^2}=50. Solve for t. You'll get 2 values:

    t_1 = 41.2 s or t_2 = 51.4 s

    During these approximately 10 s they are visible to each other.

    EB
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