An airplane flies at airspeed (relative to the
air) of 250 km/h. The pilot wishes to fly due
North (relative to the ground) but there is a
55 km/h wind blowing Southwest (direction
225◦).
In what direction should the pilot head the
plane (measured clockwise from North)?
Answer in units of ◦.
the 55 kph SW wind has a W component of magnitude ... 55 sin(45º)
the plane needs to fly N , with an E component of similar magnitude
the sine of the direction angle is ... [55 sin(45º)] / 250
Vr = Vp + 55km/h[225o] = 250,
Vp + 55*sin225 + (55*cos225)I = 250,
Vp - 38.9 - 38.9i = 250,
Vp = 88.9 + 38.9i = 97km/h[66.4o] CW.
Direction = 66.4o CW.
Correction: Last step, Vp = 288.9 + 38.9i = 292.5km/h[82.3o] CW.
Direction = 82.3o CW.
henry2
the plane flies at 250 kph
... it has to compensate for an approximate 40 kph west wind
... both of your solutions are WAY off the mark
stop confusing students
The vector sum of plane and wind should be 250 km/h due north:
Check: (288.9+38.9i) + (-38.9-38.9i) = 250 km/h.
To determine the direction in which the pilot should head the plane, we need to find the resultant velocity by considering both the airplane's airspeed and the wind velocity.
1. First, let's break down the airplane's airspeed into its northward and eastward components. We can do this using trigonometry. Let's assume the northward direction is the positive y-axis, and the eastward direction is the positive x-axis.
Given:
Airplane airspeed = 250 km/h
Angle of airplane airspeed with respect to the north = 0° (since the pilot wants to fly due north)
The northward component of the airplane's airspeed (Vy_airplane) can be calculated as:
Vy_airplane = airspeed * sin(angle)
= 250 km/h * sin(0°)
= 0 km/h
The eastward component of the airplane's airspeed (Vx_airplane) can be calculated as:
Vx_airplane = airspeed * cos(angle)
= 250 km/h * cos(0°)
= 250 km/h
Therefore, the airplane's airspeed components are Vy_airplane = 0 km/h (northward) and Vx_airplane = 250 km/h (eastward).
2. Next, let's break down the wind velocity into its northward and eastward components. We can again use trigonometry for this.
Given:
Wind speed = 55 km/h
Angle of wind direction with respect to the north = 225°
The northward component of the wind velocity (Vy_wind) can be calculated as:
Vy_wind = wind speed * sin(angle)
= 55 km/h * sin(225°)
= -38.89 km/h
The eastward component of the wind velocity (Vx_wind) can be calculated as:
Vx_wind = wind speed * cos(angle)
= 55 km/h * cos(225°)
= -38.89 km/h
Note: The negative values indicate that the wind is blowing in the opposite direction of the corresponding component.
Therefore, the wind's components are Vy_wind = -38.89 km/h (southward) and Vx_wind = -38.89 km/h (westward).
3. Now, we can find the resultant velocity by vector addition of the airplane's airspeed and the wind velocity.
Resultant northward velocity (Vy_resultant) = Vy_airplane + Vy_wind
= 0 km/h + (-38.89 km/h)
= -38.89 km/h
Resultant eastward velocity (Vx_resultant) = Vx_airplane + Vx_wind
= 250 km/h + (-38.89 km/h)
= 211.11 km/h
4. Finally, we can calculate the direction in which the pilot should head the plane. This can be done using the arctan function to find the angle between the resultant velocity vector and the positive y-axis (northward direction).
Angle = arctan(Vx_resultant / Vy_resultant)
= arctan(211.11 km/h / -38.89 km/h)
≈ -80.99°
Since the angle is negative, we can convert it to a positive angle by adding 360°.
Positive angle = 360° + (-80.99°)
≈ 279.01°
Therefore, the pilot should head the plane in the direction of approximately 279.01° clockwise from North.