(a) A jet airplane flying from Darwin, Australia, has an air...

(a) A jet airplane flying from Darwin, Australia, has an air speed of $260 \mathrm{~m} / \mathrm{s}$ in a direction $5.0^{\circ}$ south of west. It is in the jet stream, which is blowing at $35.0 \mathrm{~m} / \mathrm{s}$ in a direction $15^{\circ}$ south of east. What is the velocity of the airplane relative to the Earth? (b) Discuss whether your answers are consistent with your expectations for the effect of the wind on the plane's path.

Key Concepts

Relative Velocity

Relative velocity is the concept of describing an object’s velocity with respect to a particular frame of reference. In problems such as this, the movement of the airplane is measured relative to the moving air (jet stream) as well as the Earth. This concept requires understanding that velocities are vectors and that the observed speed and direction depend on the chosen frame of reference.

Vector Addition

Vector addition is a key concept in which two or more vectors are combined to determine a resultant vector. In the context of motion, the airplane’s airspeed vector and the wind’s velocity vector are added together to determine the airplane's actual velocity relative to the Earth. This process relies on breaking down each vector into its components and combining corresponding components.

Effect of Wind on Trajectory

The effect of wind on an airplane’s trajectory illustrates how external influences alter the intended path. Understanding this concept is important in aviation as it explains why the course over the ground may differ from the heading into the air. The wind can cause the airplane to drift off course, and pilots must adjust their heading to compensate for this effect, ensuring the airplane reaches its intended destination.

Trigonometric Decomposition

Trigonometric decomposition involves resolving vectors into perpendicular components (typically horizontal and vertical) using trigonometric functions such as sine and cosine. This decomposition is critical in accurately performing vector addition, as it allows one to handle the magnitude and directional angle for each vector separately before recombining them to form the resultant vector.

Transcript

00:01 We can say that we're going to express the two vectors in terms of their components.

00:05 So we can do unit vector.

00:06 So the velocity vector of the jet relative to the air would be equaling 260 meters per second, cosine of 5 degrees, i -hat or rather negative i -hat, plus 260 meters per second, sign of 5 degrees this would be negative j hat now the velocity vector of the air relative to the ground this would be equal to 35 meters per second cosine of 15 degrees and this would be i hat plus 35 meters per second sign of 15 degrees and this would be negative j hat and so the velocity of the jet relative to the ground would be equal to the velocity of the jet relative to the air plus the velocity of the air relative to the ground and so for the velocity of the jet relative to the ground in the x direction, this would be equal to negative 260 meters per second times cosine of 5 degrees plus 35 meters per second, cosine, other we can say, cosine of 15 degrees.

01:54 And so this is equaling negative 225 .204 meters per second.

02:03 The velocity of the jet relative to the ground in the y direction, this would be equal to negative 260 meters per second cosine of 5 degrees minus 35 meters per second cosine of 15 degrees.

02:22 And this is equal to essentially 30, negative 31 .72 meters per second.

02:48 We can say that the net magnitude, so we can say that the magnitude of the velocity of the jet relative to the ground would be equal to the square root of the sum of the squares.

03:00 So it would be 225 .204 meters per second quantity squared plus 31 .72 meters per second quantity squared...

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