Question

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?

Short answer

(a) Value of$b_x$ is - 70 cm .

(b) Value of$c_y$ is 80 cm.

(c) Magnitude of the overall displacement is 141 cm.

(d) Angle of the overall displacement relative to the x axis is counter-clockwise from the x-axis.

Step-by-step solution

Step 1: Given data

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.

Determining the concept

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 ${\overrightarrow{\mathrm{R}}}_{\mathrm{y}}$.

(a) Determination of bx

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,

$$\begin{gathered} 30.0+{\mathrm{b}}_{\mathrm{x}}-20.0-80.0=-140.0 \\ {\mathrm{b}}_{\mathrm{x}}=-70.0\mathrm{cm} \end{gathered}$$

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

(b) Determination of cy

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 $80.0\mathrm{cm}$.

(c) Determination of magnitude of overall displacement

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.

(d) Determination of angle of the overall displacement

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).