91Ó°ÊÓ

$$\begin{gathered} \left|d_1\right|=0.4 \\ \left|d_2\right|=0.5 \\ \left|d_3\right|=0.6 \end{gathered}$$

Here we need to use the concept of resolution of components and calculating resultant of given vectors. Let’s assume that positive x direction is along the east and positive y direction is along the north for the calculations.

Formulae:

$\stackrel{→}{a}+\stackrel{→}{b}+\stackrel{→}{c}=\left(a_x+b_x+c_x\right)\stackrel{→}{i}+\left(a_y+b_y+c_y\right)\stackrel{→}{j}$ (i)

$d_x=d_{1x}+d_{2x}+d_{3x}$ (ii)

$d_y=d_{1y}+d_{2y}+d_{3y}$ (iii)

$\left|d\right|=\sqrt{\left(d_x^2+d_y^2\right)}$ (iv)

When the angle is measured with respect to positive x axis, the angle made by vector ${\stackrel{→}{d}}_1$is $225°$

The x-component of vector${\stackrel{→}{d}}_1$ is calculated as,

$$\begin{gathered} {\stackrel{→}{d}}_{1x}=0.4\cos \left(225°\right) \\ =-0.2828\mathrm{m} \\ ≈-0.28\mathrm{m} \end{gathered}$$

Therefore, the x-component of vector$\stackrel{→}{d_1}$ is -0.28 m .

When the angle is measured with respect to positive x axis, the angle made by vector$\stackrel{→}{d_1}$ is $225°$

The y-component of vector$\stackrel{→}{d_1}$ is calculated as,

$$\begin{gathered} \stackrel{→}{d_{1y}}=0.4\sin (225°) \\ =-0.2828\mathrm{m} \\ ≈-0.28\mathrm{m} \end{gathered}$$

Therefore, the y-component of vector$\stackrel{→}{d_1}$ is -0.28 m.

The angle made by vector$\stackrel{→}{d_2}$with respect to positive x-axis is 0. Therefore, x component of vector$\stackrel{→}{d_2}$is,

$$\begin{gathered} \stackrel{→}{{\mathrm{d}}_{2\mathrm{x}}}=0.5\cos \left(0\right) \\ =0.50\mathrm{m} \end{gathered}$$

Therefore, the x-component of vector$\stackrel{→}{d_2}$ is 0.50 m .

The angle made by vector$\stackrel{→}{d_2}$with respect to positive x-axis is 0. Therefore, y component of vector$\stackrel{→}{d_2}$is,

localid="1661143358392" role="math" $$\begin{gathered} \stackrel{→}{{\mathrm{d}}_{2\mathrm{y}}}=0.5\sin \left(0\right) \\ =0\mathrm{m} \end{gathered}$$

Therefore, the y-component of vector $\stackrel{→}{{\mathrm{d}}_2}$is 0 .

The angle made by vector$\stackrel{→}{d_3}$ with respect to positive x-axis is$60°$ . Therefore, x component of vector$\stackrel{→}{d_3}$ is,

$$\begin{gathered} \stackrel{→}{{\mathrm{d}}_{3\mathrm{x}}}=0.6\cos (60°) \\ =0.30\mathrm{m} \end{gathered}$$

Therefore, the x-component of vector$\stackrel{→}{{\mathrm{d}}_3}$ is 0.30 m .

The angle made by vector $\stackrel{→}{{\mathrm{d}}_3}$with respect to positive x-axis is $60°$. Therefore, y component of vector $\stackrel{→}{d_3}$is,

$$\begin{gathered} \stackrel{→}{{\mathrm{d}}_{3\mathrm{y}}}=0.6\sin (60°) \\ =0.52\mathrm{m} \end{gathered}$$

Therefore, the y-component of vector$\stackrel{→}{{\mathrm{d}}_3}$ is 0.52 m .

Using the equation (ii) and parts (a), (c) and (e), the x component of ant’s net displacement is written as,

$$\begin{gathered} {\mathrm{d}}_{\mathrm{x}}={\mathrm{d}}_{1\mathrm{x}}+{\mathrm{d}}_{2\mathrm{x}}+{\mathrm{d}}_{3\mathrm{x}} \\ =0.5172\mathrm{m} \\ ≈0.52\mathrm{m} \end{gathered}$$

Therefore,the x component of ant’s net displacement is 0.52 m.

Using the equation (iii) and parts (b), (d) and (f), the y component of ant’s net displacement is written as,

$$\begin{gathered} {\mathrm{d}}_{\mathrm{y}}={\mathrm{d}}_{1\mathrm{y}}+{\mathrm{d}}_{2\mathrm{y}}+{\mathrm{d}}_{3\mathrm{y}} \\ =0.2372\mathrm{m} \\ ≈0.24\mathrm{m} \end{gathered}$$

Therefore, the y component of ant’s net displacement is 0.24 m.

Using equation (iv), calculate the magnitude of ant’s net displacement,

$$\begin{gathered} \left|\stackrel{→}{\mathrm{d}}\right|=\sqrt{\left({\mathrm{d}}_{\mathrm{x}}^2+{\mathrm{d}}_{\mathrm{y}}^2\right)} \\ =0.57\mathrm{m} \end{gathered}$$

Therefore, the magnitude of the ant’s net displacement is 0.57 m.

Direction of ant’s net displacement,

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{{\mathrm{d}}_{\mathrm{y}}}{{\mathrm{d}}_{\mathrm{x}}}\right) \\ ={\tan }^{-1}\left(\frac{0.2372}{0.5172}\right) \\ =24.7° \\ ≈25°\mathrm{north}\mathrm{of}\mathrm{east} \end{gathered}$$

Therefore, the direction of ant’s net displacement is along$25°\mathrm{north}\mathrm{of}\mathrm{east}$ .

Magnitude of ant’s net displacement when ant returns to starting point would equal to$\left|\stackrel{→}{-d}\right|$ .

Therefore, ant has to travel 0.57 m to return to its original position.

From (j) we can interpret that the direction of ant when it returns to starting point is opposite that is$25°$ south of west.