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

1. Jun 6, 2006

### VinnyCee

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 [Broken]

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!

Last edited by a moderator: May 2, 2017
2. Jun 6, 2006

### Hootenanny

Staff Emeritus
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
3. Jun 6, 2006

### VinnyCee

Thanks!

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

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

4. Jun 6, 2006

### Hootenanny

Staff Emeritus
Looks good to me

5. Jun 6, 2006

### andrevdh

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 $$V_{AB}$$.

Last edited: Jun 7, 2006