Q29E - University Physics with Modern Physics

Question

Vector $A$ has y-component $Ay = +9.60 m$. A makes an angle of $32.0$ counterclockwise from the +y-axis. (a) What is the x-component of $A$ ? (b) What is the magnitude of $A$ ?

Step-by-Step Solution

Answer

The magnitude of the vector is $21.87$ meters and the x coordinate is  $-19.66$ meter. 

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.

$= {\tan }^{-1}\left(\frac{y}{x}\right)$ is one way to express it.

Here 

$\begin{array}{l}\theta =32°\\ A_y=9.6 \mathrm{meters}\end{array}$

Step 2: Conversion of mile into kilometer

Here, 

$$\begin{gathered} \begin{array}{l}\theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right)\\ \tan \theta = \frac{A_y}{A_x}\end{array} \\ {A}_x=\frac{A_y}{\tan \theta } \end{gathered}$$

Here we will use negative sign because vector is in second quadrant

$\begin{array}{rcl}{A}_x& =& -\frac{13}{\tan \left(32°\right)}\\ & =& -19.66 \mathrm{meters}\end{array}$

Hence the x coordinate is $-19.66$meter which can be $20$ meter approx.. 

Step 3: Magnitude of the vector

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

$\begin{array}{lcl}A\begin{array}{cl}=& \sqrt{{\left(A_x\right)}^2+{\left(B_x\right)}^2}\end{array}& & \\ A=\sqrt{{\left(-19.66\right)}^2+{\left(9.6\right)}^2}& & \\ =21.87 meters& & \end{array}$

Hence the magnitude of the vector is $21.87$meters and the x coordinate is  $-19.66$ meter .