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 (y/x) is one way to express it.

Here 

$$\begin{gathered} \theta ={32}^{°} \\ {A}_y=9.6 meters \end{gathered}$$

Step 2: Conversion of mile into kilometer

Here, 

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

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

$$\begin{gathered} A_x=\frac{13}{\tan \left({32}^{°}\right)} \\ =-19.66 meters \\ Hence the x coordinate is -19.66 meter which can be 20 meter approx.. \end{gathered}$$

. 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

$$\begin{gathered} A_x=\sqrt{{\left(A_x\right)}^2 +{\left(B_x\right)}^2 } \\ =\sqrt{{\left(-19.66\right)}^2+{\left(9.6\right)}^2} \\ =21.87cmeters \end{gathered}$$

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

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