All angles are CCW.
459km/h[140o] + Vw = 498km/[156o]
Vw = 498[156] - 459[140]
X = 498*Cos156 - 459*Cos140 =-103.3 km/h
Y = 498*sin156 - 459*sin140 = -92.48km/h
Tan Ar = Y/X = -92.48/-103.3 = 0.89530
Ar = 41.84o = Reference angle.
A = 41.84 + 180 = 221.84o CCW = 41.84o
S. of W.(W41.84S).
Vw = X/Cos A = -103.3/Cos221.84 = 138.7
km/h[W41.84S].
Note: Since X and Y are both negative,
the resulting velocity is in the 3rd
Quadrant.
1. An airplane is flying with the velocity of 459km/h [E40S], however its ground velocity is 498km/h [E24S]. Determine the velocity of the blowing wind. (Use the vector subtraction diagram and geometry to solve this question).
I drew a diagram and tried solving this question by using components method.
so basically...
Vwg = 498km/h [E24S]-459km/h[E40S]
Vwg = 498km/h [E24S]+ 459km/h[W40N]
Vwg = 498km/h*cos24 +298km/h*sin24 + 459*cos40 + 459*sin40
Vwg = 455km/h[E]+203km/h[S]+351km/h[W] + 295km/h[N]
Vwg = 104km/h[E]+ 92km/h[N]
Vwg = 138.86km/h[E41.5N]
so i got 138.86km/h
1. i don't know if i am correct
2. I don't understand how to use geometry to solve the question. i drew the diagram but i don't know if that is good. I also don't get how to do vector subtraction.
Please help
thank you
1 answer