Figure 4-57 shows the path taken by a drunk skunk over level ground, from initial point i to final point f .The angles are ${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{30}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{\theta }}_{\mathbf{2}}\mathbf{=}\mathbf{50}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}$, and ${\mathit{\theta }}_3\mathbf{=}\mathbf{80}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}\mathbf{}$and the distances are ${\mathit{d}}_{\mathbf{1}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathit{m}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}{\mathit{d}}_{\mathbf{2}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathit{m}$, and ${\mathit{d}}_{\mathbf{3}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}$. What are the (a)magnitude and (b) angle of the skunk’s displacement from i to f ?

It is given that,

$$\begin{gathered} d_1=5.0m \\ {\theta }_1={30}^{°} \\ {d}_2=8.0m \\ {\theta }_2={50}^{°} \\ {d}_3=12.0m \\ {\theta }_3={80}^{°} \end{gathered}$$

This problem is based on the simple concept of magnitude and direction of resultant vector which is the sum of two or more vectors.Thus, first resolve all the displacements along x and y directions. Then, the sum of components along x and y direction can be calculated separately. Then, by using the formulas the magnitude and angle made by resultant with +x axis will be calculated.

Formula:

Magnitude of resultant vector is given by,

$R=\sqrt{R_x^2+R_y^2}$ …(i)

And the Direction of the resultant vector is given by,

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

Where, R is the resultant position vector, $R_x$is horizontal component of position vector, $R_y$is vertical component of position vector and$\theta$is direction of resultant vector.

Resultant displacement can be calculated by finding its components,

$$\begin{gathered} d_x=d_{1x}+d_{2x}+d_{3x} \\ {d}_y=d_{1y}+d_{2y}+d_{3y} \\ {\theta }_{1x}=d_1\times \cos \left({}_1\right) \\ =5.0m\times \cos \left({30}^{°}\right) \\ =4.33m \\ {\theta }_{2x}=\theta d_2\times \cos \left({180}^{°}+{}_1-{}_2\right) \\ =8.0m\times \cos \left({160}^{°}\right) \\ =-7.5m \\ {\theta }_{3x}=\theta d_3\times \theta \cos \left(360-{}_3-{}_2+{}_1\right) \\ =12.0m\times \cos \left({260}^{°}\right) \\ =-2.1m \end{gathered}$$

Hence,

$$\begin{gathered} d_x=4.33m-7.5m-2.1m \\ =-5.27m \\ {\theta }_{1y}=d_1\times \sin \left({}_1\right) \\ =5.0m\times \sin \left({30}^{°}\right) \\ =2.5m \\ {\theta }_{2y}=\theta d_2\times \sin \left({180}^{°}+{}_1-{}_2\right) \\ =8.0m\times \sin \left({160}^{°}\right) \\ =2.74m \\ {\theta }_{3y}=\theta d_3\times \theta \sin \left({360}^{°}-{}_3-{}_2+{}_1\right) \\ =12.0m\times \sin \left({260}^{°}\right) \\ =-11.82m \end{gathered}$$

Hence,

$$\begin{gathered} d_y=2.5m+2.74m-11.82m \\ =-6.58m \end{gathered}$$

Therefore, using the notation given in equation (i), the magnitude of displacement is,

$$\begin{gathered} d=\sqrt{d_x^2+d_y^2} \\ =\sqrt{\left({\left(-5.27m\right)}^2\right)+\left({\left(-6.58m\right)}^2\right)} \\ =8.43m \end{gathered}$$

Therefore, the magnitude of displacement is 8.43 m.

From the equation (ii), the direction of resultant displacement is given by

$$\begin{gathered} \theta =\left(\frac{d_y}{d_x}\right) \\ =\left(\frac{-6.58m}{-5.27m}\right) \\ ={51}^{°} \end{gathered}$$

Components of displacement are in third quadrant.

Therefore, angle from role="math" localid="1655375812403" $\mathbf{+}\mathit{x}\mathbf{}\mathit{a}\mathit{x}\mathit{i}\mathit{s}$is,

$$\begin{gathered} \theta =180+51 \\ ={231}^{°} \end{gathered}$$

If the is measured in clockwise direction the angle from $\mathbf{+}\mathit{x}\mathbf{}\mathit{a}\mathit{x}\mathit{i}\mathit{s}$is,

${\mathbf{360}}^{\mathbf{°}}\mathbf{-}{\mathbf{231}}^{\mathbf{°}}\mathbf{=}{\mathbf{129}}^{\mathbf{°}}$.

Therefore, the required angle is ${\mathbf{129}}^{\mathbf{°}}$when measured in clockwise direction.