Question

A vector velocity v has components vx = 20 m/s and vy = -31 m/s. Find the magnitude and the direction of the vector. Give the angle with respect to the horizontal positive x- axis.

Answer 1

Answer on Question #57667, Physics / Mechanics | Relativity

A vector velocity $v$ has components $v_x = 20 \, \text{m/s}$ and $v_y = -31 \, \text{m/s}$. Find the magnitude and the direction of the vector. Give the angle with respect to the horizontal positive $x$-axis.

Solution:

The magnitude is

$$|v| = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + (-31)^2} = \sqrt{1361} = 36.89$$

To find direction the following formula can be used:

$$\tan \theta = \frac{v_y}{v_x} = \frac{-31}{20} = -1.55$$ $$\theta = \tan^{-1}(-1.55) = -57.17{}^\circ = 302.83{}^\circ$$

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