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).
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
Therefore, the aircraft's ground speed is approximately 384 km/hour.
To compute the aircraft's ground speed, we need to consider the effect of the wind on the aircraft's motion. We can break down the wind vector into its North and West components.
Given:
Airspeed (Magnitude): 120 m/s
Wind speed (Magnitude): 20 m/s
Angle between wind direction and flight direction: 45 degrees
Step 1: Calculate the North and West components of the wind vector.
The North component can be calculated by taking the cosine of the given angle multiplied by the magnitude of the wind:
North component = cos(45°) x 20 m/s
The West component can be calculated by taking the sine of the given angle multiplied by the magnitude of the wind:
West component = sin(45°) x 20 m/s
Step 2: Calculate the effective North and West velocities.
The effective North velocity is the sum of the aircraft's airspeed and the North component of wind velocity:
Effective North velocity = 120 m/s + North component
The effective West velocity is the difference between the aircraft's airspeed and the West component of wind velocity:
Effective West velocity = 120 m/s - West component
Step 3: Calculate the ground speed using the Pythagorean theorem.
Ground speed is the magnitude of the resultant vector formed by the effective North and West velocities:
Ground speed = √(Effective North velocity^2 + Effective West velocity^2)
Step 4: Convert the ground speed to kilometers per hour.
To convert m/s to km/h, multiply the ground speed by 3.6:
Ground speed (km/h) = Ground speed x 3.6
Now, let's calculate the ground speed.
North component = cos(45°) x 20 m/s = 14.14 m/s
West component = sin(45°) x 20 m/s = 14.14 m/s
Effective North velocity = 120 m/s + 14.14 m/s = 134.14 m/s
Effective West velocity = 120 m/s - 14.14 m/s = 105.86 m/s
Ground speed = √(134.14^2 + 105.86^2) = 168.46 m/s
Ground speed (km/h) = 168.46 m/s x 3.6 = 606.46 km/h (rounded to two decimal places)
Therefore, the aircraft's ground speed is approximately 606.46 km/h.
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.