Question 318803: An airplane has an airspeed of 510 km/h bearing 50 deg north of east. The wind is 70 km/h in the direction 33 deg north of west. Find the resultant velocity representing the path of the airplane with resprect to the ground.
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! An airplane has an airspeed of 510 km/h bearing 50 deg north of east. The wind is 70 km/h in the direction 33 deg north of west. Find the resultant velocity representing the path of the airplane with resprect to the ground.
call North 0 degrees, East 90, South 180, West 270
50 deg north of east = 90 - 50 = 40 degreesA
33 deg north of west = 270 + 33 = 303 degrees
airplane 510 km/h at 40 degrees (call this Vector A)
wind 70 km/h at 303 degrees (call this Vector W)
okay these are vectors, vectors have a horizontal component and a vertical component and a magnitude
x = horizontal component = hypotenuse * cos(theta)
theta is the heading, the magnitude is the hypotenuse
y = vertical component = hypotenuse * sin(theta)
magnitude = sqrt(x^2 + y^2)
Vector A = 510 km/h at 40 degrees
Vector W = 70 km/h at 303 degrees
Ax = 510cos(40), Ay = 510sin(40)
Wx = 70cos(303), Wy = 70sin(303)
resultant vector horizontal component = Ax + Wx
Ax + Wx = 510cos(40) + 70cos(303) = 88.92478
resultant vector vertical component = Ay + Wy
Ay + Wy = 510sin(40) + 70sin(303)
Ay + Wy = -497.06
magnitude resultant vector = sqrt(88.92478^2 + (-497.06)^2)
magnitude resultant vector = 504.9521 km/h
direction resultant vector:
arctan of resultant vector = (Ay + Wy)/(Ax + Wx)
(Ay + Wy)/(Ax + Wx) = -497.06/88.92478 = -5.58967
arctan(-5.58967) = -79.857 degrees -->
360 - 79.857 = 280.143 degrees
the resultant velocity representing the path of the airplane with resprect to the ground is 504.95 km/h at 280.14 degrees (10.14 degrees north of west)
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