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Two boats leave shore @ different angles. How long after will they be 800ft apart?

  1. Jun 6, 2006 #1
    Two boats leave the shore at the same time and travel in the directions shown. If [itex]v_A\,=\,20\,\frac{ft}{s}[/itex] and [itex]v_B\,=\,15\,\frac{ft}{s}[/itex], determine the speed of boat A with respect to boat B. How long after leaving the shore will the boats be 800 ft apart?

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    I figured the relative velocity:

    [tex]\overrightarrow{v_A}\,=\,{\left(-20\,sin\,30\right)\,\widehat{i}\,+\,\left(20\,cos\,30\right)\,\widehat{j}}\,\frac{ft}{s}[/tex]

    [tex]\overrightarrow{v_B}\,=\,{\left(15\,cos\,45\right)\,\widehat{i}\,+\,\left(15\,sin\,45\right)\,\widehat{j}}\,\frac{ft}{s}[/tex]

    [tex]\overrightarrow{v_{A/B}}\,=\,\overrightarrow{v_B}\,-\,\overrightarrow{v_A}[/tex]

    [tex]\overrightarrow{v_{A/B}}\,=\,{(20.6)\,\widehat{i}\,+\,(-6.71)\,\widehat{j}}\,\frac{ft}{s}[/tex]

    [tex]v_{A/B}\,=\,\sqrt{20.6^2\,+\,(-6.71)^2}\,=\,21.7\,\frac{ft}{s}[/tex]

    [tex]\theta\,=\,tan^{-1}\,\right(\frac{6.71}{20.6}\right)\,=\,18[/tex]

    Here is where I can't figure out how to get the time t for when the boats are 800 ft apart. I try to complete the triangles above and solve for the length of one of the sides and get a wrong answer when I divide by the velocity to get the time since the velocity is constant. Please help!
     
  2. jcsd
  3. Jun 6, 2006 #2

    Hootenanny

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    You have found the relative velocity, i.e. how fast they are moving apart. How long will it take them to become 800m apart?
     
    Last edited: Jun 6, 2006
  4. Jun 6, 2006 #3
    Thanks!

    It seems the simpler the solution, the tougher it is to find, for me anyways!

    The right answer:

    [tex]t\,=\,\frac{800\,ft}{21.7\frac{ft}{s}}\,=\,36.9\,s[/tex]
     
  5. Jun 6, 2006 #4

    Hootenanny

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    Looks good to me :smile:
     
  6. Jun 6, 2006 #5

    andrevdh

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    The velocity of A relative to B is the apparent velocity A will have for an observer on B. It tells us how object A moves with respect to object B. That is as if object B were stationary how would object A move (with respect to it). So [tex]V_A - V_B[/tex] removes from the velocity of A the part similar to the velocity of B (their relative motion would be zero if they had similar velocities). What is left is the remaining motion of object A w.r.t. object B. So you end up with the velocity of object A with respect to object B, usually indicated as [tex]V_{AB}[/tex].
     
    Last edited: Jun 7, 2006
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