Q34 E - University Physics with Modern Physics

Question

Find the magnitude and direction of the vector represented by the following pairs of components: (a) ${\mathbf{A}}_{\mathbf{x}}\mathbf{=}\mathbf{-}\mathbf{8}\mathbf{.}\mathbf{60}\mathbf{ }\mathbf{cm}\mathbf{,}{\mathbf{A}}_{\mathbf{y}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{20}\mathbf{ }\mathbf{cm}$; (b) ${\mathbf{A}}_{\mathbf{x}}\mathbf{=}\mathbf{-}\mathbf{9}\mathbf{.}\mathbf{70}\mathbf{ }\mathbf{m}\mathbf{,}{\mathbf{A}}_{\mathbf{y}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{45}\mathbf{ }\mathbf{m}$ (c) ${\mathbf{A}}_{\mathbf{x}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{75}\mathbf{ }\mathbf{km}\mathbf{,}{\mathbf{A}}_{\mathbf{y}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{70}\mathbf{ }\mathbf{km}$  

Step-by-Step Solution

Answer

(a) the angle of the vector is $149°$ and magnitude is  10.4 cm,

(b) the angle of the vector is $14.2°$ and magnitude is 10.0m , and

(c) the angle of the vector is $109.2°$ and magnitude is  $8.21 \mathrm{km}$.

Step 1: Vector and its direction

The vector's various coordinates are presented here. We know that in coordinates, the x component of a vector comes first, followed by the y component. To calculate the angle of the vector with respect to the x-axis, just multiply the y component by the x component and then take the inverse of that result.

 

This will tell you the vector's angle with the x-axis.

 

It can be represented as 

 

$\mathbf{\theta }\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{y}{x}\right)$ …………………(1)

Step 2: (a) Magnitude and direction of A

The vector A is $(-8.60 \mathrm{cm},5.20 \mathrm{cm})$

 

The magnitude of the vector can be calculated by the sum of the square of the magnitude of the individual coordinates

$$\begin{gathered} A=\sqrt{{\left(A_x\right)}^2+{\left(B_x\right)}^2} \\ =\sqrt{(-8.6 \mathrm{cm}{)}^2+(5.2 \mathrm{cm}{)}^2} \\ =10.04 \mathrm{cm} \end{gathered}$$ 

 

 

Hence the magnitude of vector A is 10.4 cm

 

The counterclockwise angle taken from the positive x-axis is given by equation (1), such that,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right) \\ ={\tan }^{-1}\left(\frac{5.2 \mathrm{cm}}{-8.6 \mathrm{cm}}\right) \\ =\tau \tau -{\tan }^{-1}\left(\frac{5.2 \mathrm{cm}}{-8.6 \mathrm{cm}}\right) \\ =149° \end{gathered}$$

Hence vector A makes the angle of  $149°$ and magnitude is 10.4 cm.

Step 3: (b) Magnitude of the vector in case B

The vector A is $(-9.70 \mathrm{m},-2.45 \mathrm{m})$ 

 

The magnitude of the vector can be calculated by the sum of the square of the magnitude of the individual coordinates

$$\begin{gathered} A=\sqrt{{\left(A_x\right)}^2+{\left(B_x\right)}^2} \\ =\sqrt{(-9.70m{)}^2+(-2.45 \mathrm{m}{)}^2} \\ =10.0 \mathrm{m} \end{gathered}$$ 

 

Hence the magnitude of vector A is 10.4 cm 

 

The counterclockwise angle taken from the positive x-axis is given by equation (1), such that,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_{\mathrm{y}}}{A_{\mathrm{x}}}\right) \\ ={\tan }^{-1}\left(\frac{-2.45 \mathrm{m}}{-9.70 \mathrm{m}}\right) \\ =14.2° \end{gathered}$$ 

 

Hence vector A makes the angle of $194.2°$ and magnitude is 10.0 m.

Step 4: (c) Magnitude of the vector in case c

The vector A is $(-9.70\hspace{0.33em}m,\hspace{0.33em}-2.45\hspace{0.33em}m)$

 

The magnitude of the vector can be calculated by the sum of the square of the magnitude of the individual coordinates

$$\begin{gathered} A=\sqrt{{\left(A_x\right)}^2+{\left(B_x\right)}^2} \\ =\sqrt{(7.75 \mathrm{km}{)}^2+(-2.70 \mathrm{km}{)}^2} \\ =8.21 \mathrm{km} \end{gathered}$$ 

 

 

Hence the magnitude of vector A is 8.21 km  

 

The counterclockwise angle taken from the positive x-axis is given by equation (1), such that,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_{\mathrm{y}}}{A_{\mathrm{x}}}\right) \\ ={\tan }^{-1}\left(\frac{7.75 \mathrm{km}}{-2.70 \mathrm{m}}\right) \\ =\tau \tau -70.79° \\ =109.2° \end{gathered}$$ 

 

Hence vector A makes the angle of $102.9°$ and magnitude is  8.21 km.

 

Therefore for case (a) the angle of the vector is $149°$and magnitude is 10.04 cm, for case (b) the angle of the vector is $14.2°$ and magnitude is 10.0m, and in case (c) the angle of the vector is $109°$ and magnitude is 8.21 km.