Question:

$\vec{V}$ is a vector 24.8 units in magnitude and points at an angle of $23.4^\circ$ above the negative x axis.

- Sketch this vector.
- Calculate $V_x$ and $V_y$.
- Use $V_x$ and $V_y$ to obtain (again) the magnitude and direction of $\vec{V}$. [
*Note:*Part (c) is a good way to check if you've resolved your vector correctly.]

Source: Giancoli, Douglas C., Physics: Principles with Applications, 7th Edition, 2014.

Quick Answer:

- $24.8\textrm{ units}$
- $V_x = 22.8, V_y = 9.85$
- $23.4^\circ \textrm{ above the neg. x-axis}$

### Transcript for this Giancoli solution

This is Giancoli Answers with Mr. Dychko. Here's a sketch of the vector, 24.8 units long and 23.4 degree angle above the negative*x*-axis and we have to find its

*x*and

*y*components. So

*cos*of 23.4 is adjacent over hypotenuse and so that's

*v x*over

*v*and then rearrange that to solve for

*v x*by multiplying both sides by

*v*and we get this formula here. And we'll plug in the magnitude—24.8—times

*cos*of 23.4 and we get 22.8 is the

*x*-component of this vector. And likewise, for

*sin*, we use

*sin*for finding the

*y*-component of the vector. So it's gonna be 24.8 times

*sin*of 23.4 and we get 9.84 or 9.85 I guess I suppose— I should round that up maybe— for the

*y*-component of the vector. And then in part (c), we are just working backwards again and making sure we get what we started with and so as a way to check to make sure we resolved the vector correctly here and we find the magnitude by taking the sum of the squares of the components and then take the square root of that and we get 24.8 which checks out— that's what the question tells us is the length, 24.8— and then the angle is the inverse tangent of the

*y*-component divided by the

*x*-component or inverse tangent of opposite over adjacent and we get 24 degrees and according to the drawing, that's above the

*x*-axis and that checks out as well. Well, we got 23.4, we can make it have three significant figures, can't we? So 23.4 degrees above the

*x*-axis, good.