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anonymous

  • 5 years ago

Ships A and B leave port together. For the next two hours, ship A travels at 24mph in a direction 40 degree west of north while ship B travels 29 degree east of north at 32mph . what is the distance between the ships 2 hours later? What is the speed of ship A as seen by ship B?

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  1. anonymous
    • 5 years ago
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    distance between them is 48*sin40+64*sin29 m

  2. anonymous
    • 5 years ago
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    its that simple?

  3. anonymous
    • 5 years ago
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    speed of A as seen by B ((32*cos29-24*cos40)^2+(32*sin29+24*cos40)^2)^0.5

  4. anonymous
    • 5 years ago
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    so the distance is 61.88 miles and speed is 35.23 mph? right?

  5. anonymous
    • 5 years ago
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    sry it was not 4 u

  6. radar
    • 5 years ago
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    Since this form a triangle where you are given two sides and their included angle (40+29) degrees, it is also possible to use the law of cosines where\[a ^{2}=b ^{2}+c ^{2}-2bc Cos69^{o}\] This provides a value of a (distance between ships) of 64.8 why their is a discrepancy of this value between the first solution I do not know.

  7. anonymous
    • 5 years ago
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    hmm thanks! is the speed done the same way? as up there?

  8. radar
    • 5 years ago
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    I haven't looked at the speed yet, I am trying to figure the rationale, I am thinking that Jas assumed the due north directional line formed a right angle with the line between the two ships, this I believe would be an assumption and not necessarily true.

  9. radar
    • 5 years ago
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    maybe he will come back and explain his rationale! Meanwhile I ponder. lol

  10. anonymous
    • 5 years ago
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    hehe thanks.. im working on a different problem now.

  11. radar
    • 5 years ago
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    Jas, if you're back, the equation you posted for the speed ship A as observed by ship B do you intend for the square root to be taken for only the last parenthesis of for the sum of both parenthesis?

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