V = 190km/h @ 155o + 45km/h @ 15o.
X = 190*cos155 + 45*cos15=-128.73 km/h.
Y = 190*sin155 + 45*sin15=91.94 km/h.
tanAr = Y/X = 91.94/-128.73 = 0.71424
Ar = -35.53o = Reference angle.
A = -35.54 + 180 = 144.5o
V = X/cosA = -128.73/cos144.5=158.1 km/h
@ 144.5o, CCW.
Note: One student said that the book's
answer is 227 km/h @ 162.4o. Based on
the INFO given, that would be impossible
because:
1. The direction of the wind reduces
the speed of the plane.
2. The resultant direction would have to be less than 155o.
A plane is flying at 25 degrees north of west at 190km/ h. Suddenly there is a wind from 15degrees north of east at 45 km/ hr. what is the planes new velocity with respect to ground in standard position
1 answer