Calculating Wind effect

Question

An airplane is flying on a bearing of 265 degrees at 420 mph. A wind is blowing with a bearing of 210 degrees at 30 mph. Find the actual speed and direction of the plane and justify your answer.

Answer 1

draw a diagram that looks like the third quadrant

draw a line of bearing 265º from the north which represents the plane

label this 420 mph

the angle between the line and the negative x-axis is 5º

draw a line with a bearing of 210º from the north

label this 30 mph

form a parallelogram using these lines a second time or put the 30 line at the end of the 420 line

if you drew a parallelogram the diagonal is the distance you want

if you put the lines or vectors end to end and then form a triangle with a third line, this is the distance you want

use the law of cosines

let v=the distance (speed) of the plane

v²=420²+30²-(2)(420)(30)cos125º

where does the 125º come from

the 420 line is 5º from the horizontal and the 30 line is 30º from the vertical

therefore 90-30-5=55º and the adjacent angle in the parallelogram is 180-55=125º

v²=176,400+900-(25,200)(-0.57357)

v²=176,400+900+14,453.96

v²=190,853.96

v=√191,753.96

v=437.897 mph is the speed of the plane

use the law of cosines to find the direction of the plane

call the angle opposite the 30 line "x"

the law of sines says that...

30/sinx=437.897/sin125º

sinx=30sin125º/437.897

sinx=30(0.819)/437.897

sinx=24.57/437.897

sinx=0.05611

sin-¹x=0.05611

x=3.2165º

the direction of the plane is from the x-axis going counter-clockwise

the plane's vector is 5º down from the negative x-axis

you have 185º plus 3.2165º

185º+3.2165º=188.2165º is the direction of the plane

Answer 2

Angle between the wind and the airplane is 55⁰ (265 - 210).

 

Actual speed = √[(420)² + (30)² + 2.420.30.cos(55⁰)]

 

Angle with original airplane direction θ = 30 sin (55⁰) / [420 + 30 cos (55⁰)]

Direction in bearing = 265⁰ - θ