Question

Please show all work and formulas used. Thanks!

13. (Brainbuster) An airplane with a ground speed of 550 km/hr travels along a heading of 30 N of W. At that high altitude there is a powerful 210 km/hr wind (known as the jet stream) 40° S of W. Determine the direction that the pilot is aiming the plane.

Answer 1

The resultant velocity of the plane V, is given as, V: = 550 km/hr 30° N of W. The velocity of wind vis given as follows: V = 210 km/hr 40° S of W. The direction of the velocity of the plane that pilot is aiming v, is calculated as follows: Vy 550 km/hr 30° 400 210 km/hr Suppose is the angle of the vector v, from -x direction. The vector 7, can be written in the Cartesian vector form as follows: 7, --V, cos di + v, sin oj

Similarly, The vector V, can be written in the Cartesian vector form as follows: ve =-ve cos 30°î + ve sin 300 And, The vector can be written in the Cartesian vector form as follows: v=-V, cos 40º – V, sin 40° According to vector addition, V2 = 1 + -V, cos eî + v, sin ej =(-ve cos 30'î + ve sin 30°)-(-V, cos 40'î – V, sin 40°) -V, cos Oî + v, sin 0 9 =(y,cos 40°- vcos 30°) +(v, sin 40° +ve sin 30°) On equating left and right sides, -V, cos 0 = (v., cos 40º-Vcos 30º) ,cos e = v, cos 30º-V, cos 40° v, sin @ =( ,, sin 40° + vsin 30%) Now divide equation (2) by (1) and substitute 210 km hr for y, and 550 km/hr for VR- v, sine v sin 40° + v, sin 30° v, cose V, cos 30° - v,cos 40° (210 km/hr sin 40° +(550 km/hr sin 30° tan = (210 km/hr)cos 30° - (550 km/hr) cos 40° =-1.712 0 = tan- (1.712) = 59.71° *60°

Hence the pilot is aiming the plane at a direction 60° North of West.