Q 34E - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

(a) the angle of the vector is $149°$ and magnitude is 10.04 cm ,

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

(c) the angle of the vector is $109.2°$ and magnitude is  8.21 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  $\left(-8.60 \mathrm{cm}, 5.20 \mathrm{cm}\right)$

 

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

 

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

 

Hence the magnitude of vector A is  10.04 cm

 

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

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

 

Hence vector A makes the angle of $149°$ and magnitude is 10.04 cm .

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

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

 

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

 

$$\begin{gathered} \mathrm{A}=\sqrt{{\left({\mathrm{A}}_{\mathrm{x}}\right)}^2+{\left({\mathrm{B}}_{\mathrm{x}}\right)}^2} \\ =\sqrt{{\left(-9.70 \mathrm{m}\right)}^2+{\left(-2.45 \mathrm{m}\right)}^2} \\ =10.0 \mathrm{cm} \end{gathered}$$ 

 

Hence the magnitude of vector A is  10.04 cm

 

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

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

 

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  $\left(-9.70 \mathrm{m},-2.45 \mathrm{m}\right)$

 

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

 

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

 

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,

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

 

Hence vector A makes the angle of $109.2°$ and magnitude is 8.21 km .

 

Therefore for case (a) the angle of the vector is  and magnitude is  , for case (b) the angle of the vector is $14.2°$  and magnitude is 10.0 m , and in case (c) the angle of the vector is $109.2°$  and magnitude is 8.21 km .