Question

A pilot wants to fly to the North with a ground speed of 120 m/s, so he flies with an airspeed of 120 m/s. Unfortunately, he forgets about the wind: he experiences a 20 m/s wind from the North-West (so at an angle of 45 with respect to the intended heading). For this situation, compute the aircraft's ground speed (in kilometres per hour).

Answer 1

Using vector addition, we can find the resulting velocity of the aircraft in the North direction and the East direction separately.

The Northward velocity is given by:

v_N = 120 m/s * cos(45) = 84.85 m/s

The Eastward velocity is given by:

v_E = 120 m/s * sin(45) + 20 m/s = 98.88 m/s

Using the Pythagorean theorem, we can find the magnitude of the resulting velocity (ground speed):

v = sqrt(v_N^2 + v_E^2) = 129.49 m/s

Converting to kilometres per hour:

v = 129.49 m/s * 3.6 km/hour/m = 466.16 km/hour

Therefore, the aircraft's ground speed is approximately 466.16 km/hour.

Answer 2

The ground speed of the aircraft is found by adding the airspeed and the wind. Since these are not in-line, we need to do this in the x- and y-direction separately and then use Pythagoras' theorem to find the total ground speed.
Vx=Vwind * sin(45) = 20 * sin(45) = 14.14m/s
Vx = V - Vwind cos(45) = 120 = 20 * cos(45) = 105.86
Vground = sqrt(Vx^2+Vy^2) = 106.80 m/s

Now that we know our ground speed in m/s we can convert back to km/h:

106.80 m/s * 60 = 6408 m/h
6408 m/h * 60 = 384,480 km/h or 384 km/h

Answer 3

Therefore, the aircraft's ground speed is approximately 384 km/hour.