An Airplane Flies With Airspeed Relative To The Air

1 An Airplane Flies With An Airspeed Speed Relative To the Air Of 2

Identify and solve various problems related to the motion of airplanes, wind effects, displacement, resultant velocities of boats crossing rivers, and forces on sailboats, applying principles of vector addition, trigonometry, and Newton's second law.

Paper For Above instruction

Understanding the motion of objects relative to different reference frames requires a comprehensive grasp of vector addition and the physical principles underlying motion in fluid environments. The problems presented explore the influence of wind on aircraft velocity, the resultant velocity of a boat crossing a flowing river, and the acceleration resulting from forces on a sailboat. Each scenario demands different mathematical techniques, including vector decomposition, trigonometry, and Newtonian mechanics, to find solutions that reflect real-world physics.

Problem 1: Aircraft Velocity Relative to Ground with Wind Influence

An airplane's airspeed is given as 215 km/h, with the wind blowing at 45.0 km/h from various directions. When the aircraft points straight west, the ground velocity vector must be determined under different wind conditions. For each scenario, we decompose the velocity vectors into components, sum them, and find the magnitude and direction of the resultant ground velocity vector.

Given:

• Aircraft airspeed, \(V_{a} = 215\, km/h\)
• Wind speed, \(V_{w} = 45\, km/h\)
• Direction of aircraft heading: west

Case A: Wind blowing toward the west

The wind vector points toward the west; thus, it has components:

• \(V_{w_x} = -45\, km/h\)
• \(V_{w_y} = 0\)

The aircraft's velocity relative to the air points west, so its components are:

• \(V_{a_x} = -215\, km/h\)
• \(V_{a_y} = 0\)

The ground velocity components are:

• \(V_{g_x} = V_{a_x} + V_{w_x} = -215 - 45 = -260\, km/h\)
• \(V_{g_y} = 0 + 0 = 0\)

The magnitude of the ground velocity vector is:

\(V_{g} = \sqrt{V_{g_x}^2 + V_{g_y}^2} = 260\, km/h\)

The direction is due west since the y-component is zero.

Case B: Wind blowing toward the east

The wind components: \(V_{w_x} = +45\, km/h\), \(V_{w_y} = 0\)

Ground velocity components:

• \(V_{g_x} = -215 + 45 = -170\, km/h\)
• \(V_{g_y} = 0\)

Magnitude:

\(V_{g} = 170\, km/h\)

Direction remains westward but at a reduced speed due to eastward wind.

Case C: Wind blowing toward the south

The wind has components: \(V_{w_x} = 0\), \(V_{w_y} = -45\, km/h\)

Aircraft's components: \(V_{a_x} = -215\, km/h\), \(V_{a_y} = 0\)

Ground velocity components:

• \(V_{g_x} = -215 + 0 = -215\, km/h\)
• \(V_{g_y} = 0 - 45 = -45\, km/h\)

Magnitude:

\(V_{g} = \sqrt{(-215)^2 + (-45)^2} \approx \sqrt{46225 + 2025} \approx \sqrt{48250} \approx 219.6\, km/h\)

The angle relative to westward (measured south from west):

\(\theta = \arctan\left(\frac{V_{g_y}}{V_{g_x}}\right) = \arctan\left(\frac{-45}{-215}\right) = \arctan(0.209)\approx 11.8^\circ\)

The negative signs indicate the vector points slightly south of directly west, at approximately \(11.8^\circ\) south of west.

Problem 2: Aircraft Heading and Groundspeed with Wind from South

Objective: For the plane to fly directly west, determine the heading angle relative to west and the resulting groundspeed when the wind originates from the south.

• Aircraft airspeed: \(V_{a} = 215\, km/h\)
• Wind from the south at \(V_{w} = 45\, km/h\)

The wind vector points north; components: \(V_{w_x} = 0\), \(V_{w_y} = 45\, km/h\). To maintain a straight westward ground track, the aircraft must head in a direction that cancels the crosswind component.

Decomposition and Equations

Let \(\theta\) be the angle the aircraft's heading makes north of west. Components of the aircraft's velocity relative to the air:

• \(V_{a_x} = V_a \cos \theta\)
• \(V_{a_y} = V_a \sin \theta\)

Ground velocity components:

• \(V_{g_x} = V_{a_x} + V_{w_x} = V_a \cos \theta + 0\)
• \(V_{g_y} = V_{a_y} + V_{w_y} = V_a \sin \theta + 45\)

To fly straight west, \(V_{g_y} = 0\), so:

\(V_a \sin \theta + 45 = 0 \Rightarrow \sin \theta = -\frac{45}{215} \approx -0.209\)

\(\theta \approx -12.1^\circ\)

The negative angle indicates the plane points approximately \(12.1^\circ\) south of west, which is consistent with prior calculations.

The groundspeed magnitude is:

\(V_g = \sqrt{V_{g_x}^2 + V_{g_y}^2}\)

Compute \(V_{g_x}\):

\(V_{g_x} = 215 \cos(-12.1^\circ) \approx 215 \times 0.974 \approx 209.7\, km/h\)

And \(V_{g_y}\):

\(V_{g_y} = 0\, km/h\) (by the setup)

Resulting groundspeed is approximately 209.7 km/h westward.

Problem 3: Displacement and Distance of a Long Walk

A teenager walks:

• 3.0 km north
• 4.0 km east
• 5.0 km south
• 2.0 km west

Part A: Total Distance Walked

The total distance is the sum of all individual legs:

\(d_{total} = 3.0 + 4.0 + 5.0 + 2.0 = 14.0\, km\)

Part B: Displacement

Displacement is a vector from the starting point to the final point. Calculate the net north-south and east-west components:

• North-South: \(3.0\, km - 5.0\, km = -2.0\, km\) (south)
• East-West: \(4.0\, km - 2.0\, km = 2.0\, km\) (east)

The magnitude of the displacement vector:

\(D = \sqrt{(-2.0)^2 + 2.0^2} = \sqrt{4 + 4} = \sqrt{8} \approx 2.83\, km\)

The direction relative to east:

\(\phi = \arctan\left(\frac{2.0}{-2.0}\right) = \arctan(-1) = -45^\circ\)

The negative indicates the vector points southeast. The displacement magnitude is approximately 2.83 km, directed 45° south of east.

Problem 4: Velocity of a Boat Crossing a River

Given:

• Boat speed relative to water: 5.0 m/s
• River flowing east at 3.0 m/s
• Boat points north

Part A: Resultant Velocity and Angle

Boat's velocity components:

• \(V_{b_x} = 3.0\, m/s\) (east)
• \(V_{b_y} = 5.0\, m/s\) (north)

The resultant velocity vector magnitude:

\(V_{r} = \sqrt{(3.0)^2 + (5.0)^2} = \sqrt{9 + 25} = \sqrt{34} \approx 5.83\, m/s\)

The angle relative to north (measured east of north):

\(\theta = \arctan\left(\frac{3.0}{5.0}\right) \approx 30.96^\circ\)

Direction: approximately 31° east of north.

Part B: Downstream Distance

Time to cross the 100 m wide river:

\(t = \frac{distance}{V_{b_y}} = \frac{100\, m}{5.0\, m/s} = 20\, s\)

Downstream displacement during this time:

\(d = V_{b_x} \times t = 3.0\, m/s \times 20\, s = 60\, m\)

The boat will land 60 m downstream from the point directly across the river.

Problem 5: Force and Acceleration of a Sailboat

The sailboat has a mass \(m = 220\, kg\) and experiences forces:

• Force south: \(F_s = 65\, N\)
• Force west: \(F_w = 140\, N\)

Calculating Acceleration:

The net force vector components:

• \(F_x = -140\, N\) (westward)
• \(F_y = -65\, N\) (southward)

The magnitude of the net force:

\(F = \sqrt{(-140)^2 + (-65)^2} = \sqrt{19600 + 4225} = \sqrt{23825} \approx 154.4\, N\)

The angle of acceleration relative to east (or west):

\(\alpha = \arctan \left(\frac{65}{140}\right) \approx 25.9^\circ\)

Since both forces are negative in x and y directions, the acceleration points southwest at approximately \(25.9^\circ\) south of west.

The acceleration magnitude:

\(a = \frac{F}{m} \approx \frac{154.4\, N}{220\, kg} \approx 0.702\, m/s^2\)

Conclusion

The set of physics problems demonstrates the importance of vector analysis in understanding motion in the presence of external forces and environmental influences. Whether analyzing the effect of wind on an aircraft, the journey of a pedestrian, or the crossing of a river by a boat, the principles of vector addition, trigonometry, and Newtonian mechanics allow for practical predictions and insights into real-world scenarios. Mastery of these concepts is fundamental for fields ranging from aviation and maritime navigation to environmental physics and engineering applications.

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