An airplane is flying in the direction 140° with an airspeed of 450 mi/hr, and a 30 mi/hr wind is blowing in the direction 65°.

Question

An airplane is flying in the direction 140° with an airspeed of 450 mi/hr, and a 30 mi/hr wind is blowing in the direction 65°. As shown in the figure, these facts may be represented by vectors p and w of magnitudes 450 and 30, respectively. The direction of the resultant p + w gives the true course of the airplane relative to the ground, and the magnitude. ||p+w||is the ground speed of the airplane. Approximate the true course and ground speed of the airplane. (Round your answers to the nearest whole number.)

1)Group speed=

2)True course=

Answer 1

V1=450cos140°i+450sin140°j

V2=30cos65°i+30sin65°j

V1=

450*cos140=-344.72

+450*sin140= 289.25

V2=

30cos65°=12.678

+30sin65°= 27.189

V=

((-344.72i+12.678j)+(289.25i+27.189j))^.5

=(-55.47i+39.867j)^.5 OK

=-15.6^.5

≈3.95

Answer 2

Draw a sketch showing p in fourth quadrant and w in first quadrant, with “direction” assumed to be bearing which is measured clockwise from North. The effective direction of the plane is -50^o from the positive x-axis, and the wind +25^o.

Vector components <x,y> are then p = <450 cos(-50), 450 sin(-50)> = <289.3,-344.7>, and w = <30 cos 25, 30 sin 25> = <27.2,12.7>. So p + w = <316.5,-332.0>.  Group speed ||p+w|| = sqrt(x² + y²) = 459 mph. 

The angle ϑ from the positive x-axis is found from tan ϑ = y/x = (-332.0)/316.5 = -1.049, ϑ = tan^-1(-1.049) = -46^o.  Finally the bearing or True Course is 90 + 46 = 136^o.