Question

What are the horizontal and vertical components (respectively) of a 60 m/s vector at 210 degrees?1. (30'sqrt3)) m/s West, 30 m/s South 2. 60 m/s West, 30m/s South 3. 30 m/s West (30'sqrt{3}) South 4. 30 m/s West, 60 m/s South

Answer 1

Vector components: 1. (30'sqrt3)) m/s West, 30 m/s South

Explanation:

A vector can be resolved into its horizontal and vertical component by using trigonometry.

In particular, the horizontal component of a vector is given by

$v_x = v cos \theta$

where

v is the magnitude of the vector

$\theta$ is its angle measured counterclockwise from the positive x-direction

And the vertical component is given by

$v_y=v sin \theta$

In this problem, the magnitude of the vector is

$v=60 m/s$

And the angle is

$\theta=210^{\circ}$

Therefore, its  components are:

$v_x=(60)(cos 210^{\circ})=60\cdot (-\frac{\sqrt{3}}{2})=-30\sqrt{3} m/s$

$v_y=(60)(sin 210^{\circ})=60\cdot (-\frac{1}{2})=-30 m/s$

And both the components are negative: this means that the vector lies in the 3rd quadrant, so the x-component is towards west, while the y-component is towards south. So the correct answer is

1. (30'sqrt3)) m/s West, 30 m/s South

Learn more about vector components:

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