Question

The air speed indicator of a plane that took off from Detroit reads \(350 . \mathrm{km} / \mathrm{h}\) and the compass indicates that it is heading due east to Boston. A steady wind is blowing due north at \(40.0 \mathrm{~km} / \mathrm{h}\). Calculate the velocity of the plane with reference to the ground. If the pilot wishes to fly directly to Boston (due east) what must the compass read?

Short answer

Answer: The velocity of the plane with reference to the ground is approximately 353.6 km/h. The pilot should adjust the compass reading to approximately 6.49° south of due east in order to fly directly towards Boston.

Step-by-step solution

Identify the vectors

Let the airspeed vector of the plane be \(\vec{v_{p}}\) and the wind velocity vector be \(\vec{v_{w}}\). Plane's velocity with reference to the ground is the vector sum of these two vectors, denoted by \(\vec{v_{g}}\).

Break down vectors into their components

To facilitate our calculations, let us first break down the airspeed and wind velocity into their horizontal (east) and vertical (north) components. The airspeed vector is due east at \(350\, \mathrm{km/h}\), so its components are \(\vec{v_{p}} = (350.0, 0)\). The wind velocity is due north at \(40.0\, \mathrm{km/h}\), so its components are \(\vec{v_{w}} = (0, 40.0)\).

Sum the component vectors

Now, add the corresponding components of the airspeed vector, \(\vec{v_{p}}\), and the wind velocity vector, \(\vec{v_{w}}\), to find the ground velocity vector, \(\vec{v_{g}}\). East Component of \(\vec{v_{g}}\): \(V_{g_x} = V_{p_x} + V_{w_x} = 350.0 + 0 = 350.0\, \mathrm{km/h}\) North Component of \(\vec{v_{g}}\): \(V_{g_y} = V_{p_y} + V_{w_y} = 0 + 40.0 = 40.0\, \mathrm{km/h}\) Thus, the ground velocity vector \(\vec{v_{g}}\) is \((350.0, 40.0)\).

Calculate the magnitude of the ground velocity

We can now calculate the magnitude of the total ground velocity \(\vec{v_{g}}\) using the Pythagorean theorem: \(|\vec{v_{g}}| = \sqrt{V_{g_x}^2 + V_{g_y}^2} = \sqrt{350.0^2 + 40.0^2} \approx 353.6\, \mathrm{km/h}\) The velocity of the plane with reference to the ground is approximately \(353.6\, \mathrm{km/h}\).

Determine the required compass reading

We still want the ground velocity to be due east (\((350.0, 0)\)). Since the wind has a constant velocity due north, we need to adjust the plane's airspeed vector, \(\vec{v_{p'}}\), with the wind speed to achieve the desired ground velocity. So, we need to find the value of \(\vec{v_{p'}}\) such that \(\vec{v_{w}} + \vec{v_{p'}} = (350.0, 0)\). To counteract the wind's northward push, the plane must have a southward component equal in magnitude to the wind's northward component. Therefore, the new airspeed vector's components are \(\vec{v_{p'}} = (350.0, -40.0)\).

Calculate the new compass angle

The direction of the new airspeed vector \(\vec{v_{p'}}\) can be calculated using the arctangent function: Angle = \(arctan\left(\frac{V_{p'_y}}{V_{p'_x}}\right) = arctan\left(\frac{-40.0}{350.0}\right)\) Converting this angle to degrees: Angle = \(arctan\left(\frac{-40.0}{350.0}\right)*\frac{180}{\pi} \approx -6.49^\circ\) Since the angle is negative, the pilot must adjust the compass reading to approximately \(6.49^\circ\) south of due east in order to fly directly towards Boston.

Key concepts

Velocity Vectors

In physics, velocity vectors are used to represent the direction and magnitude of motion. Every velocity vector has two components: one along the horizontal axis (east-west) and one along the vertical axis (north-south). In the given exercise, the airspeed vector of the plane is due east with a magnitude of \(350 \, \text{km/h}\). Similarly, the wind has a velocity vector pointing north with a magnitude of \(40.0 \, \text{km/h}\).
These vectors can be expressed as components: the east component for the plane's airspeed vector is \(350 \, \text{km/h}\), and it has no north component, i.e., \((350.0, 0)\). For the wind, the vector is \((0, 40.0)\).
To determine the plane's velocity relative to the ground, we add these vectors together. This method, called vector addition, gives us the resultant velocity vector \(\vec{v_{g}} = (350.0, 40.0)\). This reveals the overall effect of both movements on the plane relative to the ground.

Relative Motion

Relative motion involves understanding how different objects move in relation to one another. In this exercise, we want to find the relative speed of the plane concerning the ground. The plane's airspeed (its speed in still air) and the wind's speed must be combined to give us the observed ground speed.
The horizontal movement due east by the plane is undisturbed, but the wind introduces a northward component. This requires adjustment in understanding where the plane actually heads. To find the effective velocity or 'ground velocity,' you combine both the velocities of the plane and wind using vector addition.

• The eastward velocity component remains at \(350 \, \text{km/h}\), as there's no wind resistance in that direction.
• The northward velocity is now combined to form the ground velocity, resulting in a greater overall speed than either component alone.

This concept illustrates how movements must be considered together to understand the actual path and speed of an object.

Pythagorean Theorem

The Pythagorean theorem is a key mathematical tool used to calculate the magnitude of the resultant vector when dealing with vector components. In this exercise, the plane's velocity relative to the ground, \(\vec{v_{g}}\), results from combining its eastward and northward movements.
The theorem states that the square of the length of the hypotenuse (in this case, the ground speed) is equal to the sum of the squares of the other two sides of a right triangle (the east and north components). Mathematically, it's written as:
\[|\vec{v_{g}}| = \sqrt{V_{g_x}^2 + V_{g_y}^2} = \sqrt{350.0^2 + 40.0^2} \approx 353.6\, \text{km/h}\]
This calculation allows us to find the magnitude of the plane's velocity relative to the ground, showing it's approximately \(353.6 \, \text{km/h}\), faster than the airspeed of the plane alone due east. This numerical result uses the components to present a more comprehensive view of the real-world consequence of the combined velocities.

Trigonometry in Physics

Trigonometry is often employed in physics to understand angles and directions of motion. In this particular problem, the aim is to redirect the plane's actual movement back to due east. Since the wind blows north, the pilot must adjust the plane's heading slightly south of east to counteract this drift.
To find the needed adjustment angle, the arctangent function is used, which relates the sides of a right triangle to incite directional angles:
\[\text{Angle} = \arctan\left(\frac{-40.0}{350.0}\right)\]
Converting this angle to degrees using:
\[\text{Angle} = \arctan\left(\frac{-40.0}{350.0}\right)*\frac{180}{\pi} \approx -6.49^\circ\]
This calculation signifies that the pilot must adjust the plane's heading \(6.49^\circ\) south of east to stay on course to Boston.

• The negative sign indicates the need to head south.
• Understanding these adjustments ensures precise navigation despite environmental factors like wind.

By using trigonometry, pilots can effectively compute necessary course changes to maintain desired flight paths.