Boats A & B: Speed & Separation

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Two boats leave the shore simultaneously, traveling at speeds of 20 ft/s and 15 ft/s in different directions. The relative velocity of boat A with respect to boat B is calculated to be approximately 21.7 ft/s. To determine when the boats will be 800 ft apart, the time is calculated using the formula t = distance/relative velocity, resulting in 36.9 seconds. The discussion highlights the importance of understanding relative motion to find the apparent velocity of one object from the perspective of another. The solution emphasizes that simplifying the problem can sometimes lead to confusion, but the correct approach yields the desired result.
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Two boats leave the shore at the same time and travel in the directions shown. If v_A\,=\,20\,\frac{ft}{s} and v_B\,=\,15\,\frac{ft}{s}, determine the speed of boat A with respect to boat B. How long after leaving the shore will the boats be 800 ft apart?

http://img366.imageshack.us/img366/8164/engr204problem121995cj.jpg

I figured the relative velocity:

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

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

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

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

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

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

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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You have found the relative velocity, i.e. how fast they are moving apart. How long will it take them to become 800m apart?
 
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Thanks!

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

The right answer:

t\,=\,\frac{800\,ft}{21.7\frac{ft}{s}}\,=\,36.9\,s
 
VinnyCee said:
Thanks!

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

The right answer:

t\,=\,\frac{800\,ft}{21.7\frac{ft}{s}}\,=\,36.9\,s

Looks good to me :smile:
 
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 V_A - V_B 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 V_{AB}.
 
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