Question

A ship leaves the island of Guam and sails 285km at 40.0 degrees north of west. In which direction must it now head and how far must it sail so that its resultant displacement will be 115km directly east of Guam

Answer 1

Let $G$ is Guam, $A$ is the point of the first stop and $B$ is the destination.

Let we have 3 vectors: $\vec{u} = \overrightarrow{AG}$ , $\vec{v} = \overrightarrow{AB}$ and $\vec{w} = \overrightarrow{GB}$ .

Then we have the following vector equation:

$$\vec {u} + \vec {v} = \vec {w}$$

Let $\vec{v} = (x,y)$ . Then for projections we have the following 2 equations:

For $x: -285 * \cos 40{}^\circ + x = 115$ .

For $y: 285 * \sin 40{}^\circ + y = 0$ .

Solving these equations we get:

$$x = 115 + 285 * \cos 40{}^\circ \approx 333$$ $$y = -285 * \sin 40{}^\circ \approx -183$$

Let's find the distance $AB$ and the angle:

$$| \vec {v} | = \sqrt {x ^ {2} + y ^ {2}} = \sqrt {333^ {2} + 183^ {2}} = \sqrt {110889 + 33489} \approx 380 \mathrm {km}.$$

Angle: $A = \operatorname{atan}\left|\frac{y}{x}\right| = \operatorname{atan}\left| - \frac{183}{333}\right| \approx 29{}^\circ$ .

So, the ship must sail $380\mathrm{km}$ at 29 degrees south of east.