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Chapter 3: Problem 64

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

Expert verified
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

01

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 \(\vec{v_{g}}\).
02

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 \(\vec{v_{w}} = (0, 40.0)\).
03

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 \((350.0, 40.0)\).
04

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 \(353.6\, \mathrm{km/h}\).
05

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 \(\vec{v_{p'}} = (350.0, -40.0)\).
06

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.

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