Vector V1 is 6.29 units long and points along the negative x axis. Vector V2 is 4.23 units long and points at +35.0° to the +x axis.
(a) what is V1x & V1y
(b) what is the magnitude & direction (degrees)
you can do this. You have the x,y component of them added.
magnitude=sqrt(x^2+y^2)
angle=arctan y/x
but whats my X & Y
V1x=-6.29 V2y=0
V2x=4.23cos35 V2y=4.23Sin35
Do you need a tutor? Surely there are example problems in the text. This is very basic problems.
for some reason the 6.29 for vx1 keeps telling me i'm wrong..thanks a lot for the help! i obviously suck at this!
it is in the NEGATIVE Direction.
To find the components of vector V1, which points along the negative x-axis, you can use the fact that the magnitude (length) of V1 is given as 6.29 units. Since V1 is pointing along the negative x-axis, its y-component is 0, and its x-component will be negative.
(a) V1x = -6.29
V1y = 0
To find the magnitude and direction (angle) of V2, you can use the length given as 4.23 units and the angle given as +35.0° to the positive x-axis.
(b) Magnitude of V2 = 4.23 units
Direction of V2 = +35.0° to the +x axis
Note that the angle is measured counterclockwise from the positive x-axis.