A Physics question with Trig...

• Feb 10th 2010, 05:28 PM
M23
A Physics question with Trig...
Hi

I'm stumped on a physics question here, which goes:

The vector, v, has the following components: Vx= +65.0 m/s , Vy= -120 m/s

Calculate the magnitude and polar positive direction of the vector, v.

(Sorry, but please assume the "v" as the v with an arrow over it, and the x and y in Vx and Vy as subscripts...)

So, using that information, I graphed out a triangle and found the hypotenuse.

http://imgur.com/rnpjg.jpg

To get the polar positive direction, I believe I need the degree of the angle that the red arrow is pointing to. But how do I find that angle?

Thanks a lot.

(Also, am I right in thinking that the hypotenuse is the magnitude of the vector?)
• Feb 10th 2010, 05:37 PM
skeeter
Quote:

Originally Posted by M23
Hi

I'm stumped on a physics question here, which goes:

The vector, v, has the following components: Vx= +65.0 m/s , Vy= -120 m/s

Calculate the magnitude and polar positive direction of the vector, v.

(Sorry, but please assume the "v" as the v with an arrow over it, and the x and y in Vx and Vy as subscripts...)

So, using that information, I graphed out a triangle and found the hypotenuse.

http://imgur.com/rnpjg.jpg

To get the polar positive direction, I believe I need the degree of the angle that the red arrow is pointing to. But how do I find that angle?

Thanks a lot.

(Also, am I right in thinking that the hypotenuse is the magnitude of the vector?)

yes, $|v| = \sqrt{(v_x)^2 + (v_y)^2}$

for an angle in quad IV ...

$\theta = 360^\circ + \arctan\left(\frac{v_y}{v_x}\right)$
• Feb 10th 2010, 09:43 PM
M23
Thank you so much. I was trying the same thing, but it was not working, until I realized my calculator was in radian mode :(

Thanks anyways for confirming!