# Thread: A Physics question with Trig...

1. ## 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.

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?)

2. 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.

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

for an angle in quad IV ...

$\displaystyle \theta = 360^\circ + \arctan\left(\frac{v_y}{v_x}\right)$

3. 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!