Question

Consider a particle with initial velocity v⃗ that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis. 1.What is the x component vx of v⃗ ? 2.What is the y component vy of v⃗ ?

Answer 1

Answer:

-6.0 m/s, 10.4 m/s

Explanation:

To find the x- and y- components, we have to apply the formulas:

$v_x = v cos \theta$

$v_y = v sin \theta$

where

v = 12.0 m/s is the magnitude of the vector

$\theta$ is the angle between the direction of the vector and the positive x-axis

Here, the angle given is the angle above the negative x-axis; this means that the angle with respect to the positive x-axis is

$\theta=180^{\circ} - 60^{\circ} = 120^{\circ}$

So, the two components are:

$v_x = (12.0 m/s) cos 120^{\circ}=-6.0 m/s$

$v_y = (12.0 m/s) sin 120^{\circ}=10.4 m/s$