Q30E - University Physics with Modern Physics

Question

Vector A  is in the direction 34.0 clockwise from the -yaxis. The x-component of A is Ax = -16.0 m. (a) What is the y-component of A ? (b) What is the magnitude of A ?

Step-by-Step Solution

Answer

the magnitude of the vector is 28.6 meters and the x coordinate is  23.7 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.

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

Here 

 

 = 90-34 º=56 (Clockwise)

A_x = 16 meter 

=tan-1yx

Step 2: Finding out direction

Here, 

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

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

A_y = 16 tan 56

= 23.7 meter

Hence the y coordinate is 23.7 meter which can be 24 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

$$\begin{gathered} A=\sqrt{{\left(A_x\right)}^2 +{\left(B_x\right)}^2} \\ =\sqrt{{\left(23.7\right)}^2+{\left(16\right)}^2} \\ =28.6 meter \end{gathered}$$

Hence the magnitude of the vector is 28.6 meters and the x coordinate is  23