Consider a particle with initial velocity that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis.
What is the x component Vx of V?
What is the y component Vy of V?

- anonymous

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- Hunus

|dw:1329692762479:dw|
The x component Vx of V is the velocity vector V projected onto the x axis.
We obtain this magnitude as follows
\[v_{x}=12\cos(120)=-12\cos(60)=-12(\frac{1}{2})=-6\]
The y component Vy of V is the velocity vector V projected onto the y axis.
We obtain this magnitude as follows
\[v_{y}=12\sin(120)=12\sin(60)=12\frac{\sqrt{3}}{2}=6\sqrt{3}=10.4\]
|dw:1329693129338:dw|

- anonymous

why does sin have a positive value?

- Hunus

y has a positive value in the top two quadrants

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## More answers

- Hunus

|dw:1329693437661:dw|

- anonymous

so to figure out positive and negative i have to look at 12

- Hunus

You have to look at the direction it's pointing in. For theta equal to 120 or '60 above the negative x axis' the vector is pointing in the negative x direction and in the positive y direction

- anonymous

so what is u have something like this :

- anonymous

srry i mean

- anonymous

|dw:1329693963427:dw|

- anonymous

now its moving clockwise

- Hunus

The vector A you described is at an angle of (180+60) 240 degrees with respect to the positive x axis as you can see from your picture it is directed toward the negative x negative y direction, so both x and y will be negative.

- anonymous

why do i have to add 180?

- Hunus

|dw:1329694073557:dw|
Because we're making the angle with respect to the positive x axis to make it simple to take the sine and cosine of the vector

- Hunus

By adding 180 to 60 we're changing it from your original picture
|dw:1329694180018:dw|
and making it
|dw:1329694218137:dw|

- anonymous

or u can say the angle is -60

- Hunus

If we're taking it to be with respect to the positive x axis
|dw:1329694331261:dw|
then it would be -120 degrees.
-60 degrees looks like this
|dw:1329694414386:dw|

- anonymous

|dw:1329694584288:dw|

- anonymous

thats how u show it

- anonymous

cause its moving clockwise

- Hunus

Yes.

- anonymous

so in this question is the 60 degrees going clockwise

- Hunus

It typically doesn't matter how you arrive at the angle, but it appears that the angle was rotated counter clockwise 60 degrees. The vector is at 240 degrees with respect to the positive x axis.

- anonymous

so qhen do i know when to add 180 or minus 180

- Hunus

Look at the positive x axis and see how many degrees it takes to get to your vector.
This is only showing whether a vectors components are positive of negative.

- Hunus

or negative*

- anonymous

can u please give me an example of each type if itsnt too much trouble

- Hunus

|dw:1329695075597:dw|
Is vector M's x component positive or negative? What about its y component?

- anonymous

the x compoinent is negative while the y component is positive

- Hunus

\[v_x=|M|\cos(150)=|M|(\frac{-\sqrt{3}}{2})=\frac{-|M|\sqrt{3}}{2}\]so we know that the value of Vx is negative while the value of Vy is positive.
From this we could say that
\[v_{y}=|M| \sin(30)=\frac{|M|}{2}\]
and
\[v_{x}=-|M|\cos(30)=-|M|\frac{\sqrt{3}}{2}\]
Where |M| is the magnitude of the vector M
However we could also say that, to get to vector M from the positive x axis
|dw:1329695464157:dw|
we would have to go through an angle of 150 degrees.
Thus would make our calculations
\[v_y=|M|\sin(150)=\frac{|M|}{2}\]
As you can see we arrive at the same answer which is why it doesn't matter which way you get to the angle.

- Hunus

\[v_{x}=|M|\cos(150)=|M|(\frac{-\sqrt{3}}{2})=\frac{-|M|\sqrt{3}}{2}\]

- anonymous

what does 150 degrees tellus ?

- anonymous

are we saying the angle 150 is equal to 30 degrees

- Hunus

150 degrees tells us what the angle is from the positive x axis. That's what we usually like to use as a reference point when using sines and cosines.

- Hunus

What we are saying is that we only need the reference angle (how many degrees it is from the closest x axis) if we know if x is positive or negative or y is positive or negative

- anonymous

so basically we are saying the angle is 150 degrees away from the orgin

- Hunus

From the positive x axis

- anonymous

what about the negative x axis

- Hunus

150 degrees tells us how far we would have to rotate a vector from the positive x axis (0 degrees) to get to where the vector is.

- anonymous

ohhh i get it

- Hunus

Good :)

- anonymous

and to find the negative value what would i have to do

- Hunus

Well, the degrees from the positive x axis tells us whether it's positive or negative. For example, between 0 and 90 degrees, x and y are negative, between 90 and 180 degrees x is negative and y is positive, between 180 and 270 degrees both x and y are negative and from 270 to 360 degrees x is positive and y is negative. If you, for example, put in 6sin(340) you will get a negative value because y is negative in that quadrant. If you put in 12cos(320) you will get a positive value because x is positive in that quadrant.

- Hunus

between 0 and 90 x and y are positive*** sorry

- anonymous

so it doesnt matter whether u say 30 degreees or 150 degrees

- anonymous

thanks . could u please help me with another question

- Hunus

It matters, but only as far as the dirrection goes. This is because the MAGNITUDE of cos(30) is the same as the MAGNITUDE of cos(150), but one is negative and one is positive

- Hunus

Sorry for the delay on that response

- anonymous

and that has to do withe quadrant whether is positive or negative

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