91影视

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:

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

$\mathrm{R}=\left|\stackrel{鈫�}{\mathrm{R}}\right|=鈭�\sqrt{({\stackrel{鈫�}{\mathrm{R}}}_{\mathrm{x}}^2+{\stackrel{鈫�}{\mathrm{R}}}_{\mathrm{y}}^2}\left(\mathrm{ii}\right)$

$\mathrm{胃}={\tan }^{-1}\frac{\stackrel{鈫�}{{\mathrm{R}}_{\mathrm{y}}}}{\stackrel{鈫�}{{\mathrm{R}}_{\mathrm{x}}}}\left(\mathrm{iii}\right)$

Where,

${\stackrel{鈫�}{\mathrm{R}}}_{\mathrm{x}},{\stackrel{鈫�}{\mathrm{R}}}_{\mathrm{y}},\stackrel{鈫�}{\mathrm{R}}$are radius vectors and $\mathrm{胃}$is the angle between${\stackrel{鈫�}{R}}_x$ and ${\stackrel{鈫�}{\mathrm{R}}}_{\mathrm{y}}$.

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 $(-140.0鈥�\mathrm{m},-20.0鈥�\mathrm{m})$.

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 $(-140.0鈥�\mathrm{m},-20.0鈥�\mathrm{m}).$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,

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

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

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