physics
posted by shantel on .
A pilot flies her route in two straightline segments. The displacement vector A for the first segment has a magnitude of 246 km and a direction 30.0o north of east. The displacement vector B for the second segment has a magnitude of 178 km and a direction due west. The resultant displacement vector is R = A + B and makes an angle è with the direction due east. Using the component method, find (a) the magnitude of R and (b) the directional angle è.

a. R = 240km @ 30o+178km @ 180o
X = 240*cos30 + ()178 = 29.8 km.
Y = 240*sin30 = 120 km.
R^2 = X^2 + Y^2 = (29.8)^2+(120)^2=15288
R = 123.6 km.
tanA = Y/X = 120/29.8 = 4.03.
A = 76.1o North of East = Direction 
Correction: Change 240 km to 246 km and redo all calculations.