Question

Rewrite the following vectors in terms of their magnitude and angle (counterclockwise from the +x direction).


1) A displacement vector with an x component of +9.5 m and a y component of -9.6 m.
magnitude:


2) angle:



3) A velocity vector with an x component of -96 m/s and a y component of +46 m/s.
magnitude:



4) angle:



5) A force vector with a magnitude of 58 lb that is in the third quadrant with an x component whose magnitude is 37 lb.
y component:


6) angle:

Answer 1

1. Disp. = X - yi -= 9.5 - 9.6i =
Sqrt(9.5^2+9.6^2) =

2. Tan A = (-9.6/9.5 = -1.01053.
A = -45.3o,Q4 = 45o S. of E. = 315o CCW.

3. V = -96 + 46i = Sqrt(96^2+46^2) =

4. Tan A = 46/(-96) = -0.47917.
A = -25.6o, Q2 = 25.6o N. of W. = 154.4o CCW.

Answer 2

5. X^2 + Y^2 = 58^2.
37^2 + Y^2 = 58^2. Y = ?.

6. Q3, -37 - Yi.
Calculate the angle.