Question

A pilot flies her airplane from its initial position to a position \(200.0 \mathrm{~km}\) north of that spot. The airplane flies with a speed of \(250.0 \mathrm{~km} / \mathrm{h}\) with respect to the air. The wind is blowing west to east with a speed of \(45.0 \mathrm{~km} / \mathrm{h}\). In what direction should the pilot steer her plane to accomplish this trip? (Express your answer in degrees noting that east is \(90^{\circ},\) south is \(180^{\circ},\) west is \(270^{\circ}\), and north is \(360^{\circ} .\) )

Short answer

In this problem, we are given the airplane's speed (250 km/h) and the wind's speed (45 km/h, blowing west to east). We need to find the direction the pilot should steer her plane to reach a point 200 km north of the initial position. After performing the necessary calculations and analyzing the velocity vectors, we find that the pilot should steer her plane towards 283.29 degrees (east of north).

Step-by-step solution

Determine the airplane's velocity vector with respect to the ground

Let's first represent the airplane's velocity with respect to the air as a vector. Since the airplane is flying at a speed of \(250.0 \mathrm{~km} / \mathrm{h}\), the magnitude of its velocity vector is \(250.0 \mathrm{~km} / \mathrm{h}\). Let's denote the direction angle of this vector as \(\theta\). Now, we can represent the airplane's velocity vector as: $$\vec{v}_a = 250.0(\cos\theta \hat{i} + \sin\theta \hat{j}) \mathrm{~km} / \mathrm{h}$$

Determine the wind's velocity vector

The wind is blowing west to east with a speed of \(45.0 \mathrm{~km} / \mathrm{h}\). This means that its velocity vector point in the eastward direction, and we can represent it as: $$\vec{v}_w = 45.0 \hat{i} \mathrm{~km} / \mathrm{h}$$

Subtract the wind's velocity vector from the airplane's velocity vector

To find the required velocity vector \(\vec{v}_r\) for the airplane to have a ground velocity pointing exactly north, we need to subtract the wind's velocity vector from the airplane's ground velocity vector: $$\vec{v}_r = \vec{v}_a - \vec{v}_w$$ Now substituting the vectors and solving for \(\vec{v}_r\), we get: $$\vec{v}_r = 250.0(\cos\theta \hat{i} + \sin\theta \hat{j}) - 45.0 \hat{i} \mathrm{~km} / \mathrm{h}$$ Since we want the airplane to have a ground velocity pointing north, the horizontal component of \(\vec{v}_r\) should be equal to zero: $$250.0\cos\theta - 45.0 = 0$$ Solving for \(\cos\theta\): $$\cos\theta = \frac{45}{250}$$

Use the arctangent function to find the angle corresponding to the required velocity vector

Now, we will use the arctangent function to find the angle \(\theta\). Since we know \(\cos\theta\), we can find \(\sin\theta\) using the Pythagorean identity: $$\sin\theta = \sqrt{1 - \cos^2\theta}$$ $$\sin\theta = \sqrt{1 - \left(\frac{45}{250}\right)^2}$$ Now we can find the angle \(\theta\) using the arctangent function: $$\theta = \tan^{-1}\left(\frac{\sin\theta}{\cos\theta}\right)$$ $$\theta = \tan^{-1}\left(\frac{\sqrt{1 - \left(\frac{45}{250}\right)^2}}{\frac{45}{250}}\right)$$ Calculating the value of \(\theta\) in degrees: $$\theta \approx 76.71^{\circ}$$ Since north is represented as \(360^{\circ}\), the direction the pilot should steer her plane is \(\theta\) degrees east of north. The pilot should steer her plane towards \(\boxed{360^{\circ} - 76.71^{\circ} = 283.29^{\circ}}\).

Key concepts

Vector Subtraction

Vector subtraction is a fundamental concept in physics, especially in problems involving motion. When you have multiple forces or velocities acting on an object, such as an airplane, you need to combine these vectors to understand the resultant motion.
Consider the velocity of an airplane and the velocity of the wind as vectors. The airplane's velocity with respect to the ground (its ground velocity) can be determined by subtracting the wind's velocity vector from the airplane's velocity vector. This is because the wind's speed can either aid or impede the airplane's motion relative to the ground.

In vector terms, if \(\vec{v}_a\) is the airplane's velocity with respect to the air, and \(\vec{v}_w\) is the wind's velocity, then the airplane's ground velocity \(\vec{v}_r\) is given by \(\vec{v}_r = \vec{v}_a - \vec{v}_w\). To perform this subtraction visually, you can place the tail of the wind's velocity vector at the head of the airplane's air velocity vector, and draw the resultant vector from the tail of the airplane's vector to the head of the wind's vector.

Airplane Ground Velocity

The airplane's ground velocity is its velocity relative to a fixed point on the Earth. It is crucial to understand that this ground velocity is not just the speed at which the airplane is moving through the air; it also includes the effect of the wind.
To find the airplane's ground velocity vector, we must account for both its velocity through the air and the wind velocity. The pilot needs to adjust the airplane's air velocity by steering into the wind at an angle that will counteract the wind's effect. This will result in a straight path towards the intended direction when observed from the ground.

The ground velocity is the vector sum of the wind velocity and the velocity of the airplane through the air. By adjusting the control surfaces, the pilot can change the airplane's heading to achieve the desired ground path.

Wind Velocity Vector

The wind velocity vector represents the speed and direction of the wind's influence on any object moving through the air. In aviation, understanding and accounting for the wind's velocity are essential for accurate navigation.

For example, in our exercise, the wind has a constant velocity of \(45.0 \mathrm{~km} / \mathrm{h}\) from west to east, which can be represented as a vector pointing in the positive x-direction (east) on a coordinate plane.

To visualize this, if you draw a horizontal arrow pointing right, you have a representation of the wind velocity vector with a magnitude (length) proportional to the wind's speed. Pilots factor this wind velocity into their course to ensure they reach their destination accurately, as failure to do so would result in being blown off course.

Direction Angle Calculation

The direction angle calculation is the process of determining the angle a vector makes with a reference direction, such as North. In aviation, this reference is typically true North, and angles are measured clockwise from North.
To calculate the direction angle that the pilot must steer the airplane, we use trigonometry. In our example, the angle \(\theta\) required for the airplane to counteract the wind and travel north is found using the arctangent function, which gives us the angle between the x-axis and the vector when the vector's components along the x and y axes are known.

The pilot must steer \(\theta\) degrees east of north to maintain a northern course. This is because the wind is adding a component to the airplane's velocity in the eastern direction, so the pilot must adjust her course westward by the proper angle. Knowing the correct direction angle to steer is crucial for maintaining the intended flight path despite environmental conditions like wind.