Question

An airplane flies due north with an air speed of260 km/h . A steady wind at 65 km/h blows eastward. (Air speed is the speed relative to the air.)

What is the plane’s ground speed (*v*pg)?

If the pilot wants to fly due north, what should his heading be?

Answer #1

In this the given the speed of the airplane relative to air is 260km/h duenorth

therefore v_{pa} = 260km/h due north

then v_{pa,x} = 0km/h

v_{pa,y} = 260km/h

also given speed of wind relative to ground is 65km/s dueeast.

therefore....the solution are....

v_{ag} = 65km/h due east

then v_{ag,x} = 65km/h

v_{ag,y} = 0km/h

Then

v_{pg,x} = v_{pa,x} +v_{ag,x}

Substitute the value.....

= 0km/h + 65km/h

= 65km/h

And v_{pg,y} = v_{pa,y} +v_{ag,y}

=260km/h + 0km/h

= 260km/h

Then

v_{pa} = sqrt[(65km/h)^{2}
+(260km/h)^{2}]

=268km/h

direction is tan^{-1}(260/65) = 75.9^{o}toward
north of east

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