Question

What are the x- and y- components of a vector that has a 42.9 m magnitude and a direction of 252

Answer 1

-13.3 m, -40.8 m

Step-by-step explanation:

We can find the components of the vector by using the equations:

`v_x = v cos \theta`

`v_y = v sin \theta`

where

v is the magnitude of the vector

`\theta` is the angle representing the direction of the vector (measured above the positive x-direction)

For the vector in this problem,

v = 42.9 m

`\theta=252^(\circ)`

Therefore its components are

`v_x = (42.9) ( cos 252^(\circ))=-13.3 m\\v_y = (42.9)( sin 252^(\circ))=-40.8 m`