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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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Most popular questions from this chapter

In an arcade game, a ball is launched from the corner of a smooth inclined plane. The inclined plane makes a \(30.0^{\circ}\) angle with the horizontal and has a width of \(w=50.0 \mathrm{~cm}\) The spring-loaded launcher makes an angle of \(45.0^{\circ}\) with the lower edge of the inclined plane. The goal is to get the ball into a small hole at the opposite corner of the inclined plane. With what initial velocity should you launch the ball to achieve this goal?

In ideal projectile motion, when the positive \(y\) -axis is chosen to be vertically upward, the \(y\) -component of the velocity of the object during the ascending part of the motion and the \(y\) -component of the velocity during the descending part of the motion are, respectively, a) positive, negative. c) positive, positive. b) negative, positive. d) negative, negative.

What is the maximum height above ground a projectile of mass \(0.79 \mathrm{~kg}\), launched from ground level, can achieve if you are able to give it an initial speed of \(80.3 \mathrm{~m} / \mathrm{s} ?\)

You serve a tennis ball from a height of \(1.8 \mathrm{~m}\) above the ground. The ball leaves your racket with a speed of \(18.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(7.00^{\circ}\) above the horizontal. The horizontal distance from the court's baseline to the net is \(11.83 \mathrm{~m},\) and the net is \(1.07 \mathrm{~m}\) high. Neglect spin imparted on the ball as well as air resistance effects. Does the ball clear the net? If yes, by how much? If not, by how much did it miss?

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