91Ó°ÊÓ

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.

• 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.

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

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,

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

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,

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

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,

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

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,

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

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,

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

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,

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

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,

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

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,

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

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,

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

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,

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

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,

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

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,

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

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,

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

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,

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

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,

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

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,

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

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,

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

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

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

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