Relative motion of two ships

  • Thread starter quick
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  • #1
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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.
 
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Answers and Replies

  • #2
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anyone help please? im stuck
 
  • #3
Pyrrhus
Homework Helper
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try using

[tex] v = \frac{d}{t} [/tex]
 
  • #4
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i got it using the law of cosines. thanks for your suggestion though.
 
  • #5
NateTG
Science Advisor
Homework Helper
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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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