Q34E - University Physics with Modern Physics

Question

Find the magnitude and direction of the vector represented by the following pairs of components: (a) $A_x=-8.60 \mathrm{cm},A_y=5.20 \mathrm{cm};$ (b) $A_{\mathrm{x}}=-9.70 \mathrm{m},A_{\mathrm{y}}=-2.45 \mathrm{m}$(c) $A_{\mathrm{x}}=7.75 \mathrm{km},A_{\mathrm{y}}=-2.70 \mathrm{km}$

Step-by-Step Solution

Answer

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

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

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

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

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

(a) Magnitude and direction of A

The vector $\mathrm{A}$ 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 $10.04 \mathrm{cm}$

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

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

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

(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 $10.04 \mathrm{cm}$

The counterclockwise angle taken from the positive $\mathrm{x}$-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 $10.0 \mathrm{m}$.

(c) Magnitude of the vector in case c

The vector $\mathrm{A}$ 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

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

Hence the magnitude of vector A is $8.21 \mathrm{km}$

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

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

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

Therefore for case (a) the angle of the vector is $149°$ and magnitude is $10.04 \mathrm{cm}$, for case (b) the angle of the vector is $14.2°$and magnitude is $10.0 \mathrm{m}$, and in case (c) the angle of the vector is $109.2°$ and magnitude is $8.21 \mathrm{km}$.