91Ó°ÊÓ

1)$\stackrel{→}{\mathrm{P}}=10.0 \mathrm{m} \mathrm{at} 25.0°\mathrm{counterclockwise}\mathrm{from} +\mathrm{x}$

2)$\stackrel{→}{\mathrm{Q}}=12.0 \mathrm{m} \mathrm{at} 10.0°\mathrm{counterclockwise}\mathrm{from} +\mathrm{y}$

3)$\stackrel{→}{\mathrm{R}}=8.00 \mathrm{m} \mathrm{at} 20.0°\mathrm{clockwise}\mathrm{from} -\mathrm{y}$

4)$\stackrel{→}{\mathrm{S}}=9.00 \mathrm{m}\mathrm{at} 40.0°\mathrm{counterclockwise}\mathrm{from}-\mathrm{y}$

We can use vector addition law and find the net displacement of given vectors. By using the trigonometry formula, we find its direction and magnitude.

Formula:

$\stackrel{→}{d}=\stackrel{→}{P}+\stackrel{→}{Q}+\stackrel{→}{R}+\stackrel{→}{S}$ i)

$d=\sqrt{{\left(d_x\right)}^2+{\left(d_y\right)}^2}$ (ii)$\tan \mathrm{θ}=\frac{d_y}{d_x}$ (iii)

The vector configuration can be drawn as below:

The sum of the four vectors is,

$\stackrel{→}{d}=\stackrel{→}{P}+\stackrel{→}{Q}+\stackrel{→}{R}+\stackrel{→}{S}$

The x component of four vectors is,

$\stackrel{→}{d_x}={\stackrel{→}{P}}_x+\stackrel{→}{Q_x}+\stackrel{→}{R_x}+\stackrel{→}{S_x}$

Substitute the value of each component from the figure of vector configuration.

$$\begin{gathered} {\stackrel{→}{\mathrm{d}}}_{\mathrm{x}}=\left(\mathrm{Pcos}25-\mathrm{Qsin}10-\mathrm{Rsin}20+\mathrm{Ssin}40\right)\stackrel{Áåœ}{\mathrm{i}} \\ =\left(10.0\cos 25-12.0\sin 10-8.00\sin 20+9.00\sin 40\right)\stackrel{Áåœ}{\mathrm{i}} \\ =10.03\stackrel{Áåœ}{\mathrm{i}} \end{gathered}$$

Similarly, the y component of four vectors is

$\stackrel{→}{d_y}={\stackrel{→}{P}}_y+\stackrel{→}{Q_y}+\stackrel{→}{R_y}+\stackrel{→}{S_y}$

Substitute the value of each component from the figure of vector configuration.

$$\begin{gathered} {\stackrel{→}{\mathrm{d}}}_{\mathrm{y}}=\left(Ps\mathrm{in}25+Q\cos 10-R\cos 20-S\cos 40\right)\stackrel{Áåœ}{j} \\ =\left(10\sin 25+12\cos 10-8\cos 20-9\cos 40\right)\stackrel{Áåœ}{j} \\ =1.6319\stackrel{Áåœ}{j} \end{gathered}$$

In unit vector notation,

$$\begin{gathered} \stackrel{→}{d}=\stackrel{→}{d_x\stackrel{Áåœ}{i}}+\stackrel{→}{d_y}\stackrel{Áåœ}{j} \\ =10.03\stackrel{Áåœ}{i}+1.63\stackrel{Áåœ}{j} \end{gathered}$$

Therefore, the sum of four vectors,$\stackrel{→}{d}$ is equal to $10.03\stackrel{Áåœ}{i}+1.63\stackrel{Áåœ}{j}$.

Use the equation (ii) to calculate the magnitude of vector$\stackrel{→}{d}$.

$$\begin{gathered} \mathrm{d}=\sqrt{{\left({\mathrm{d}}_{\mathrm{x}}\right)}^2+{\left({\mathrm{d}}_{\mathrm{y}}\right)}^2} \\ =\sqrt{{\left(10.03\right)}^3+{\left(1.6319\right)}^2} \\ =10.2\mathrm{m} \end{gathered}$$

Therefore, the magnitude of$\stackrel{→}{d}$ is 10.2m.

Now, calculate the direction using equation (iii) and the x and y components of the vector $\stackrel{→}{d}$.Therefore, the direction is,

$$\begin{gathered} \tan \mathrm{θ}=\frac{d_y}{d_x} \\ \mathrm{θ}={\tan }^{-1}\left(\frac{d_y}{d_x}\right) \\ ={\tan }^{-1}\left(\frac{1.63}{10.03}\right) \\ =9.24° \end{gathered}$$

Therefore, the angle of the vector$\stackrel{→}{d}$ is$9.24°$ in counterclockwise direction from the positive x-axis.