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?(adsbygoogle = window.adsbygoogle || []).push({});

http://img366.imageshack.us/img366/8164/engr204problem121995cj.jpg [Broken]

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!

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

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