An ant crazed by the sun on a hot Texas afternoon, darts over aplane scratched in the dirt. The and component of four executive darts are the following. All in centimetres:$\mathbf{(}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{,}\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}{\mathbf{b}}_{\mathbf{x}}\mathbf{-}\mathbf{70}\mathbf{.}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{-}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{,}{\mathbf{c}}_{\mathbf{y}}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{-}\mathbf{80}\mathbf{.}\mathbf{0}\mathbf{,}\mathbf{-}\mathbf{70}\mathbf{.}\mathbf{0}\mathbf{)}\mathbf{.}$The overall displacement of the four darts has thecomponents. What are a)and b)? What are c) magnitude and d) angle (relative to the positive direction of theaxis) of the overall displacement?

For consecutive darts, the x and y components are as below:,

$(30.0,40.0),({\mathrm{b}}_{\mathrm{x}},-70),(-20.0,{\mathrm{c}}_{\mathrm{y}}),(-80.0,-70.0),$

The coordinates of the final displacement vectoris localid="1657021510505" $(-140.0,-20.0).$

All the coordinates are in centimetres.

Adding vectors in the same direction and equation that to a component of the resultant vector in the same direction. From these missing values of the component can be calculated.

Formulae are as follows:

$\mathrm{a}⃗+\mathrm{b}⃗+\mathrm{c}⃗+\mathrm{d}⃗=({\mathrm{a}}_{\mathrm{x}}+{\mathrm{b}}_{\mathrm{x}}+{\mathrm{c}}_{\mathrm{x}}+{\mathrm{d}}_{\mathrm{x}})\hat{\mathrm{i}}+({\mathrm{a}}_{\mathrm{y}}+{\mathrm{b}}_{\mathrm{y}}+{\mathrm{c}}_{\mathrm{y}}+{\mathrm{d}}_{\mathrm{y}})\left(\mathrm{i}\right)$

$\mathrm{R}=\left|\overrightarrow{\mathrm{R}}\right|=\surd \sqrt{({\overrightarrow{\mathrm{R}}}_{\mathrm{x}}^2+{\overrightarrow{\mathrm{R}}}_{\mathrm{y}}^2}\left(\mathrm{ii}\right)$

$\mathrm{\theta }={\tan }^{-1}\frac{\overrightarrow{{\mathrm{R}}_{\mathrm{y}}}}{\overrightarrow{{\mathrm{R}}_{\mathrm{x}}}}\left(\mathrm{iii}\right)$

Where,

${\overrightarrow{\mathrm{R}}}_{\mathrm{x}},{\overrightarrow{\mathrm{R}}}_{\mathrm{y}},\overrightarrow{\mathrm{R}}$are radius vectors and $\mathrm{\theta }$is the angle between${\overrightarrow{R}}_x$ and ${\overrightarrow{\mathrm{R}}}_{\mathrm{y}}$.

To find,${\mathrm{b}}_{\mathrm{x}}$all components of all vectors are needed to be added which is equal to -140.0 because components of resultants are $(-140.0\hspace{0.33em}\mathrm{m},-20.0\hspace{0.33em}\mathrm{m})$.

For each vector along x-axis,

Therefore, the value ${\mathrm{b}}_{\mathrm{x}}$of - 70.0 cm .

The sum of all y components is equal to -20.0 ,because components of resultant are $(-140.0\hspace{0.33em}\mathrm{m},-20.0\hspace{0.33em}\mathrm{m}).$Therefore,

$$\begin{gathered} 40.0-70.0+{\mathrm{c}}_{\mathrm{y}}-70.0=-20.0 \\ {\mathrm{c}}_{\mathrm{y}}=80.0\mathrm{cm} \end{gathered}$$

Hence,

the value ${\mathrm{c}}_{\mathrm{y}}$of $80.0\mathrm{cm}$.

The magnitude of the resultant displacement$(-140,-20.0)$is,

$\sqrt{(-140{)}^2+(-20.0{)}^2}=141\mathrm{cm}$

Therefore, the magnitude of the overall displacement is 141 cm.

An overall displacement is in the third quadrant, the angle of the overall displacement is given by,

$\mathrm{\pi }+{\tan }^{-1}\left(\frac{-20.0}{-140}\right)=188°$

Therefore, the angle of the overall displacement is ${188}^{°}$counter clockwise from x axis.

The angle can also be written as$360°-188=172°$, in the clockwise direction from x-axis (or$-{172}^{°}$ in clockwise direction).