Question

In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: $\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{\text{km}}$$\mathbf{45}\mathbf{.}\mathbf{0}\mathbf{°}$north of west; then $\mathbf{4}\mathbf{.}\mathbf{70}\mathbf{}\mathbf{\text{km}}$$\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{°}$south of east; then $\mathbf{1}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{\text{km}}$$\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{°}$south of west; then $\mathbf{5}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{\text{km}}$straight east; then $\mathbf{1}\mathbf{.}\mathbf{70}\mathbf{}\mathbf{\text{km}}$$\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{°}$east of north; then $\mathbf{7}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{\text{km}}$ $\mathbf{55}\mathbf{.}\mathbf{0}\mathbf{°}$south of west; and finally $\mathbf{2}\mathbf{.}\mathbf{80}\mathbf{\text{km}}$$\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{°}$north of east. What is his final position relative to the island?

Short answer

The Gilligan is located at a distance $7.34\mathrm{km}$, $63.5^{°}$ south of east of his island.

Step-by-step solution

Displacement

Displacement can be measured with the help of the final and initial positions of the object that is in motion.

The shortest distance between any two points is a straight line.

Given data

• The magnitude of vector $A$is, $A=2.5\text{km}$.
• The magnitude of vector $\mathit{B}$is, $B=4.7\text{km}$.
• The magnitude of vector $C$ is, $C=1.3\text{km}$ .
• The magnitude of vector $D$ is, $D=5.1\text{km}$.
• The magnitude of vector $E$ is, $E=1.7\text{km}$.
• The magnitude of vector $F$is, $F=7.2\text{km}$.
• The magnitude of vector G is, $G=2.8\text{km}$.
• The direction of the vector $A$ is $45°$ north of west.
• The direction of the vector $B$is $60°$south of east.
• The direction of the vector$C$ is $25°$ west of south.
• The direction of the vector $D$ is straight east.
• The direction of the vector $E$ is $5.0°$east of north.
• The direction of the vector $F$ is $55°$south of west.
• The direction of the vector$G$ is $10°$ north of east.

Vector representation

The vector $A$,$B$,$C$,$D$,$E$,$F$and $G$is represented as,

Horizontal component of the vectors

The horizontal component of the vector $A$is,

$A_x=-A\cos \left(45°\right)$

Here $A$is the magnitude of the vector $A$.

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}{A}_x& =& -\left(2.5\text{km}\right)\times \cos \left(45°\right)\\ & =& -1.77\text{km}\\ & & \end{array}$

The horizontal component of the vector $B$is,

$B_x=B\cos \left(60°\right)$

Here $B$is the magnitude of vector $B$.

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}{B}_x& =& \left(4.7\text{km}\right)\times \cos \left(60°\right)\\ & =& 2.35\text{km}\\ & & \end{array}$

The horizontal component of the vector $C$ is,

$C_x=-C\cos \left(25°\right)$

Here$C$is the magnitude of vector$C$.

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}{C}_x& =& -\left(\text{1.3}\text{km}\right)\times \cos \left(25°\right)\\ & =& -1.18\text{km}\\ & & \end{array}$

The horizontal component of the vector $D$is,

$D_x=D\cos \left(0°\right)$

Here$D$is the magnitude of vector$D$.

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}{D}_x& =& \left(\text{5.1}\text{km}\right)\times \cos \left(0°\right)\\ & =& 5.1\text{km}\\ & & \end{array}$

The horizontal component of the vector $E$ is,

$E_x=E\sin \left(5°\right)$

Here $E$is the magnitude of vector $E$.

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}{E}_x& =& \left(\text{1.7}\text{km}\right)\times \sin \left(5°\right)\\ & =& 0.15\text{km}\\ & & \end{array}$

The horizontal component of the vector $F$is,

$F_x=-F\cos \left(55°\right)$

Here $F$is the magnitude of vector$F$.

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}{F}_x& =& -\left(\text{7.2}\text{km}\right)\times \cos \left(55°\right)\\ & =& -4.13\text{km}\\ & & \end{array}$

The horizontal component of the vector$G$is,

$G_x=G\cos \left(10°\right)$

Here$G$is the magnitude of vector $G$.

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}{G}_x& =& \left(\text{2.8}\text{km}\right)\times \cos \left(10°\right)\\ & =& 2.76\text{km}\\ & & \end{array}$

The horizontal component of the resultant vector $R$ is,

$R_x=A_x+B_x+C_x+D_x+E_x+F_x+G_x$

Substitute the values in the above expression, and we get,

$$\begin{gathered} \begin{array}{rcl}{R}_x& =& \left\{\left(-1.77\text{km}\right)+\left(2.35\text{km}\right)+\left(-1.18\text{km}\right)+\left(5.1\text{km}\right) \\ +\left(\text{0.15}\text{km}\right)+\left(-4.13\text{km}\right)+\left(2.76\text{km}\right) \\ \right\}\\ & =& 3.28\text{km}\\ & & \end{array} \end{gathered}$$

Vertical components of vectors A, B

The vertical component of the vector $A$is,

$A_y=A\sin \left(45°\right)$

Here $A$is the magnitude of vector $A$.

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}{A}_y& =& \left(2.5\text{km}\right)\times \sin \left(45°\right)\\ & =& 1.77\text{km}\\ & & \end{array}$

The vertical component of the vector $B$ is,

$B_y=-B\sin \left(60°\right)$

Here $B$ is the magnitude of vector $B$.

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}{B}_y& =& -\left(4.7\text{km}\right)\times \sin \left(60°\right)\\ & =& -4.07\text{km}\\ & & \end{array}$

Vertical components of vectors C, D

The vertical component of the vector $C$is,

$C_y=-C\sin \left(25°\right)$

Here $C$is the magnitude of vector $C$.

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}{C}_y& =& -\left(\text{1.3}\text{km}\right)\times \sin \left(25°\right)\\ & =& -0.55\text{km}\\ & & \end{array}$

The vertical component of the vector D is,

$D_y=D\sin \left(0°\right)$

Here$D$ is the magnitude of vector $D$.

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}{D}_y& =& \left(\text{5.1}\text{km}\right)\times \sin \left(0°\right)\\ & =& 0\\ & & \end{array}$

Vertical components of vector E, F

The vertical component of the vector $E$ is,

$E_y=E\cos \left(5°\right)$

Here $E$ is the magnitude of vector $E$.

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}{E}_y& =& \left(\text{1.7}\text{km}\right)\times \cos \left(5°\right)\\ & =& 1.69\text{km}\\ & & \end{array}$

The vertical component of the vector $F$ is,

$F_y=-F\sin \left(55°\right)$

Here $F$is the magnitude of vector $F$.

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}{F}_y& =& -\left(\text{7.2}\text{km}\right)\times \sin \left(55°\right)\\ & =& -5.9\text{km}\\ & & \end{array}$

Vertical components of vector G, H

The vertical component of the vector $G$ is,

$G_y=G\sin \left(10°\right)$

Here$G$ is the magnitude of vector $G$.

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}{G}_y& =& \left(\text{2.8}\text{km}\right)\times \sin \left(10°\right)\\ & =& 0.49\text{km}\\ & & \end{array}$

The vertical component of the resultant vector $R$ is,

$R_y=A_y+B_y+C_y+D_y+E_y+F_y+G_y$

Substitute the values in the above expression, and we get,

$$\begin{gathered} \begin{array}{rcl}{R}_y& =& \left\{\left(1.77\text{km}\right)+\left(-4.07\text{km}\right)+\left(-0.55\text{km}\right)+\left(0\right)\backslash \mathrm{hfill} \\ +\left(1.69\text{km}\right)+\left(-5.9\text{km}\right)+\left(0.49\text{km}\right)\backslash \mathrm{hfill} \\ \right\}\\ & =& -6.57\text{km}\\ & & \end{array} \end{gathered}$$

Magnitude and direction of the resultant vector

The magnitude of the resultant vector $R$ is,

$R=\sqrt{R_x^2+R_y^2}$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}R& =& \sqrt{{\left(3.28\text{km}\right)}^2+{\left(-6.57\text{km}\right)}^2}\\ & =& 7.34\text{km}\\ & & \end{array}$

The direction of the resultant vector is,

$\theta ={\tan }^{-1}\left(\frac{R_y}{R_x}\right)$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}\theta & =& {\tan }^{-1}\left(\frac{-6.57\text{km}}{3.28\text{km}}\right)\\ & =& -63.5°\\ & & \end{array}$

Hence, the Gilligan is located at a distance of $7.34\mathrm{km}$, $63.5°$south of east from his island.