91影视

In a two-dimensional coordinate system, vectors can be split into component x and component y. Using these components, the magnitude of the vector is calculated. Also, the magnitude angle notation can be used to find out the direction of the vector. Here, east is indicating positive x axis and north is indicating positive y axis.

Given

Here vector $\stackrel{鈫�}{d_1}$has magnitude $d_1=4.0\mathrm{m}$. Similarly

$\stackrel{鈫�}{d_2}$has magnitude $d_2=5.00m$and

$\stackrel{鈫�}{d_2}$has magnitude $d_3=6.0\mathrm{m}$

Using the standard notation of direction and taking all the angles positive with respect to positive x axis in the anticlockwise direction, it gives

$$\begin{gathered} {胃}_1=225掳\mathrm{for}{\mathrm{d}}_1 \\ {\mathrm{胃}}_2=0\mathrm{for}{\mathrm{d}}_2\mathrm{and} \\ {\mathrm{胃}}_3=60掳\mathrm{for}{\mathrm{d}}_3 \\ \mathrm{The}\mathrm{component}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{vector}\mathrm{is}\mathrm{given}\mathrm{by} \\ {\mathrm{d}}_{\mathrm{x}}=\mathrm{d}\mathrm{肠辞蝉胃}\left(\mathrm{i}\right) \\ {\mathrm{d}}_{\mathrm{y}}=\mathrm{d}\mathrm{蝉颈苍胃}\left(\mathrm{ii}\right) \end{gathered}$$

Substituting the given magnitude and direction in equation (i) and (ii),

The x and y component of $d_1$is written as

$$\begin{gathered} d_{1x}=d_1\cos {胃}_1=-2.83\mathrm{m} \\ {\mathrm{d}}_{1\mathrm{y}}={\mathrm{d}}_1{\mathrm{蝉颈苍胃}}_1=-2.83\mathrm{m} \end{gathered}$$

Substituting the given magnitude and direction in equation (i) and (ii),

The x and y component of $d_2$is written as

$$\begin{gathered} d_{2x}=d_2\cos {胃}_1=5\mathrm{m} \\ {\mathrm{d}}_{2\mathrm{y}}={\mathrm{d}}_2{\mathrm{蝉颈苍胃}}_1=0\mathrm{m} \end{gathered}$$

Substituting the given magnitude and direction in equation (i) and (ii),

The x and y component of$d_3$ is written as

$$\begin{gathered} d_{3x}=d_3\cos {胃}_3=3.0\mathrm{m} \\ {\mathrm{d}}_{3\mathrm{y}}={\mathrm{d}}_3{\mathrm{蝉颈苍胃}}_3=5.20\mathrm{m} \end{gathered}$$

The net displacement for x and y component is given by

$$\begin{gathered} d_x=d_{1x}+d_{2x}+d_{3x} \\ {d}_x=-2.83\mathrm{m}+5.0+3.0\mathrm{m} \\ {\mathrm{d}}_{\mathrm{x}}=5.17\mathrm{m} \\ {\mathrm{d}}_{\mathrm{x}}={\mathrm{d}}_{1\mathrm{y}}+{\mathrm{d}}_{2\mathrm{y}}+{\mathrm{d}}_{3\mathrm{y}} \\ {\mathrm{d}}_{\mathrm{x}}=-2.83\mathrm{m}+0\mathrm{m}+5.20\mathrm{m} \\ {\mathrm{d}}_{\mathrm{x}}=2.37\mathrm{m} \end{gathered}$$

The magnitude of net displacement is given by

$$\begin{gathered} d=\sqrt{d_x^2+d_y^2} \\ d=\sqrt{{\left(5.17m\right)}^2+{\left(2.37m\right)}^2} \\ d=5.69m \end{gathered}$$

The direction is given by

$$\begin{gathered} 胃={\tan }^{-1}\left(\frac{d_y}{d_x}\right)={\tan }^{-1}\left(\frac{2.37}{5.17}\right) \\ 胃=24.6掳 \end{gathered}$$

This means it pointing at $25掳$north of east

When the particle returns to its initial position, the net displacement is 0. In this case

For the particle to return to its original position, the new displacement should be negative or opposite to its net displacement. Therefore, the new displacement has same magnitude of 5.69 m. And it is pointing in opposite direction so, the direction is at $25掳$south of west.