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nath
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?
|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|
why does sin have a positive value?
y has a positive value in the top two quadrants
so to figure out positive and negative i have to look at 12
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
so what is u have something like this :
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.
|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
By adding 180 to 60 we're changing it from your original picture |dw:1329694180018:dw| and making it |dw:1329694218137:dw|
or u can say the angle is -60
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|
cause its moving clockwise
so in this question is the 60 degrees going clockwise
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.
so qhen do i know when to add 180 or minus 180
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.
can u please give me an example of each type if itsnt too much trouble
|dw:1329695075597:dw| Is vector M's x component positive or negative? What about its y component?
the x compoinent is negative while the y component is positive
\[v_x=|M|\cos(150)=|M|(\frac{-\sqrt{3}}{2})=\frac{-|M|\sqrt{3}}{2}\]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.
\[v_{x}=|M|\cos(150)=|M|(\frac{-\sqrt{3}}{2})=\frac{-|M|\sqrt{3}}{2}\]
what does 150 degrees tellus ?
are we saying the angle 150 is equal to 30 degrees
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.
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
so basically we are saying the angle is 150 degrees away from the orgin
what about the negative x axis
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.
and to find the negative value what would i have to do
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.
between 0 and 90 x and y are positive*** sorry
so it doesnt matter whether u say 30 degreees or 150 degrees
thanks . could u please help me with another question
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
Sorry for the delay on that response
and that has to do withe quadrant whether is positive or negative