A plane has an airspeed of 127 km/h. It is flying on a bearing of 78 degrees while there is a 21 km/h wind out of the northeast (bearing 225 degree). What are the ground speed and the bearing of the plane?



Answer :

Answer:

  110 km/h on a bearing of 84°

Step-by-step explanation:

You want the sum of the vectors 127 km/h at 78° and 21 km/h at 225°.

Vector sum

The vector sum is most easily found using a calculator that can do math with numbers in polar coordinates. The result is shown in the attachment.

The plane's ground speed and bearing are 110 km/h at 84°.

The hard way

Each speed can be decomposed into a speed to the north and a speed to the east. The (north, east) vector components are ...

  (north, east) = speed·(cos(bearing), sin(bearing))

  127∠78° = (127)(cos(78°), sin(78°)) ≈ (26.405, 124.225)
  21∠225° = (21)(cos(225°), sin(225°)) ≈ (-14.849, -14.849)

Then the sum of these vectors is ...

  ground speed = (26.405 -14.849, 124.225 -14.849) = (11.556, 109.376)

The speed is the Pythagorean sum of these speeds:

  speed = √(11.556² +109.376²) ≈ 109.985 . . . . km/h

And the bearing angle is ...

  bearing = arctan(east/north) = arctan(109.376/11.556) ≈ 83.97°

The (north, east) speeds are both positive indicating the direction is in the first quadrant. No adjustment of the angle is needed to account for the quadrant.

The plane's ground speed and bearing are 110 km/h at 84°.

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