Vectors Magnitudes and Angles (Respect to the X Axis)

[FAJISTE]

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.

Homework Equations

The Attempt at a Solution

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

a) Answer is 23.75
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!

 

[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 :(

 

[prime-factor]

Hmmm. What answer is given (assuming an answer is given)?

 

[rl.bhat]

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 want to get it wrong!

 

[rl.bhat]

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.