Q28E - University Physics with Modern Physics

Question

Let   be the angle that the vector   makes with the +x-axis, measured counterclockwise from that axis. Find angle   for a vector that has these components: (a) $A_x=2.00m,A_y=-1.00m$ ; (b$A_x=2.00m,A_y=1.00m$ ; (c) $A_x=-2.00m,A_y=1.00m$ ; (d) $A_x=-2.00m,A_y=-1.00m$  

Step-by-Step Solution

Answer

The angle made by the vectors can be summarized as, 

(a) $333.4^o$ ,

(b)$26.6^o$  ,

(c) $333.4^o$ , and

(d) $26.6^o$ .

Angle between two vectors

Here different coordinates of the vector are given. We are knowing that in coordinates $x$ component of the vector will be first and then the $y$ component. Now to find the angle of the vector with respect to the $x$-axis we just need to take the ratio of the $y$ component to that of the $x$ component and then take the tan inverse of that value.

This will give you the angle of the vector with the x-axis.

It can be represented as

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

(a)Calculation of the θ for A x = 2 . 00 m ,   A y = 1 . 00 m

Substitute the given data in equation (1), and we get,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_{\mathrm{x}}}\right) \\ ={\tan }^{-1}\left(\frac{-1.00 \mathrm{m}}{2.00 \mathrm{m}}\right) \\ ={\tan }^{-1}\left(\frac{-1}{2}\right) \\ =-26.6° \\ =360°-26.6° \\ =333.4° \end{gathered}$$

(b) Calculation of the θ for A x = 2 . 00   m , A y = 1 . 00   m

Substitute the given data in equation (1), and we get,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right) \\ ={\tan }^{-1}\left(\frac{1.00 \mathrm{m}}{2.00 \mathrm{m}}\right) \\ ={\tan }^{-1}\left(\frac{1}{2}\right) \\ =26.6° \end{gathered}$$

(c) Calculation of the θ for A x = - 2 . 00   m , A y = 1 . 00   m

Substitute the given data in equation (1), and we get,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_{\mathrm{x}}}\right) \\ ={\tan }^{-1}\left(\frac{-1.00 \mathrm{m}}{2.00 \mathrm{m}}\right) \\ ={\tan }^{-1}\left(\frac{-1}{2}\right) \\ =-26.6° \\ =360°-26.6° \\ =333.4° \end{gathered}$$

(d) Calculation of the θ for A x = - 2 . 00   m , A y = - 1 . 00   m

Substitute the given data in equation (1), and we get,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_{\mathrm{x}}}\right) \\ ={\tan }^{-1}\left(\frac{-1.00 \mathrm{m}}{-2.00 \mathrm{m}}\right) \\ ={\tan }^{-1}\left(\frac{1}{2}\right) \\ =-26.6° \end{gathered}$$

Therefore the angle made by the vectors is (a)$333.4^o$,(b)$26.6^o$,(c)$333.4^o$,(d)$26.6^o$