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Homework Help: Relative motion of two ships

  1. Sep 28, 2004 #1
    Ships A and B leave port together. For the next two hours, ship A travels at 35.0 mph in a direction 25.0 degrees west of north while the ship B travels 80.0 degrees east of north at 30.0 mph.
    What is the distance between the two ships two hours after they depart?
    What is the speed of ship A as seen by ship B?

    basically i tried using trig to find the x component of both vectors and then added them. it wanted distance at t = 2 so i did 70 * sin (25) + 60 * sin(80) but it was incorrect. i tried this method with the example in the book and it was fairly close. so im guessing i did it incorrectly but got an answer that was close by coincidence? and for the second part im really unsure where to begin, i just know reference frames have something to do with it and maybe some sort of vector addition/subtraction. any suggestions will be much appreciated.
     
    Last edited: Sep 28, 2004
  2. jcsd
  3. Sep 28, 2004 #2
    anyone help please? im stuck
     
  4. Sep 28, 2004 #3

    Pyrrhus

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    try using

    [tex] v = \frac{d}{t} [/tex]
     
  5. Sep 28, 2004 #4
    i got it using the law of cosines. thanks for your suggestion though.
     
  6. Sep 28, 2004 #5

    NateTG

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    When you're calculating the lenght of a component-wise vector it's
    [tex]\sqrt{x^2+y^2}[/tex]
    from the way you describe what you did, you might have forgotten the [tex]y[/tex] component.

    The second part looks like straightforward vector addition as well.
     
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