Q 30E - 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, we  multiply the y component by the x component and then take the inverse of that result.

This tells us the vector's angle with the x axis.

= tan^-1 (y/x) is one way to express it.

Here, 

 

= 90-34 º=56 (Clockwise)

A_x = 16 meter 

Step 2: Finding out the direction

Here, 

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

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

A_y = 16 tan 56

= 23.7 meter

Hence, the y coordinate is 23.7 meter which can be rounded off to 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.

$A=\sqrt{{\left(A_x\right)}^2+}{\left(B_x\right)}^2$

= $\sqrt{{\left(23.7\right)}^2+{\left(16\right)}^2}$

=28.6  meter 

 

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