Question

14a. A vector A has components A_x = −2.00 m and A_y = 2.50 m.

 Find the magnitude (in m) and the direction (in degrees counterclockwise from the +x-axis) of the vector.

Answer 1

$$A=\sqrt{A_x^2+A_y^2}=\sqrt{(-2.00)^2+2.50^2}=3.20\:\rm m$$

$$\theta=\tan^{-1}\frac{A_y}{A_x}=\tan^{-1}\frac{2.50}{-2.00}=129^\circ$$