Physics problems dealing with Unit Vectors way too early in the morning...?

Hello, how does one solve these problems dealing with vectors???








1. A particle undergoes a displacement r of magnitude 46 m in a direction 25° below the x axis. Express the change in r in terms of the unit vectors x and y.


___mX + ____mY








2. If A = (23 m) x + (-12 m) y, and B = (2.0 m) x + (14 m) y, find the direction and magnitude of each of the following vectors.


(a) A


____° (counterclockwise from the +x-axis)


____m





(b) B


____° (counterclockwise from the +x-axis)


____m





(c) A + B


____° (counterclockwise from the +x-axis)


____m








Thank you so much for any help given.

Physics problems dealing with Unit Vectors way too early in the morning...?
1.


46cos25 mx + -46sin25 my





2.





TanΘ = y/x


magnitude = √x²+y²


a)


TanΘ = 12/23


Θ = 27.55281158 below the x-axis .


Counter clockwise = 360 - Θ = 332.4471884


magnitude = √x²+y² = 25.94224354


b)


TanΘ = 14/2


Θ = 81.86989765 from +x-axis


Counterclocwise = 360 - Θ = 278.130124


magnitude = √(2)² + (14)² = 14.14213562





c)


A = 23mx -12my


B = 2mx + 14my


A+B = 25mx + 2my





TanΘ = 2/25


Θ = 4.57392126


counterclockwise = 360 - Θ = 355.4260787





magnitude = √(25)²+(2)² = 25.07987241 N
Reply:Here are the formulas to remember:





X=Rcos(Angle)


Y=Rsin(Angle)





The R stands for the hypotenous, or total displacement, in this case 46m for question one. The angle has to be in the RCS angle, which means measured counter-clockwise from the Positive X Axis. In question one, that would mean 335*.





So, for question one, It's:





X=Rcos(Angle)


X=(46m)cos(335*)


X=41.69m





Y=Rsin(Angle)


Y=(46m)sin(335*)


Y=-19.44m





2. a) -27.55*


25.94m





b) 81.87*


14.14





c) 4.57*


25.07m