# Vectors Magnitudes and Angles (Respect to the X Axis)

#### FAJISTE

1. Homework Statement

a) If Vx = 18.00 units and Vy = -15.50 units, determine the magnitude of V.
b) Determine the angle of V with respect to the x-axis.

2. Homework Equations

3. The Attempt at a Solution

Yeah, Part B is pretty much the only one I don't get .

b) I'm not too sure about what the whole "respect to the x-axis" means. The angle does lie in the fourth quadrant, so I assumed the answer is to do 360 degrees - the angle (i got 40.73) and get 319.27 but that didn't work. I'm down to one try for the answers, so I have to get this one right.

Thanks!

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#### prime-factor

a) If Vx = 18.00 units and Vy = -15.50 units, determine the magnitude of V.

Magnitude = sqrt ( x^2 + y^2)

=> Magnitude = sqrt ( [(18).(18)] + [(-15.5)(-15.5)] )
=> Magnitude = sqrt (324 + 240.25)
=> Magnitude = sqrt ( 564.25)
=> Magnitude = 23.75

With part B I get the same answer as you using:

tan^-1 (y /x)

tan^-1 (15.5/18)

=> Angle = 40.73 degrees

#### FAJISTE

Yeah, that answer is incorrect. I don't get what it means by respect to the x axis

#### prime-factor

If you want to measure it with respect to the x axis it means that you start on the x - axis and move counter clockwise to your angle. I assume?

#### FAJISTE

Ya, that gives me 360 - 40.73 = 319.27, which was wrong :(

#### rl.bhat

Homework Helper
Wright the angle as -40.73 degree. Because angle in clockwise is taken as negative.

#### FAJISTE

Are you sure? I have one last try and I don't wanna get it wrong!

Homework Helper
I think so.

#### FAJISTE

im scared to try iot :(

#### FAJISTE

can anyone confirm -40.73?

#### jerrdan

Yes, -40.73 is what I get too.

This means the angle is 40.73 degrees below the horizontal axis.