Problem 71 - University Physics with Modern Physics

Question

An airplane pilot sets a compass course due west and maintains an airspeed of 220 km/h. After flying for 0.500 h, she finds herself over a town 120 km west and 20 km south of her starting point. (a) Find the wind velocity (magnitude and direction). (b) If the wind velocity is 40 km/h due south, in what direction should the pilot set her course to travel due west? Use the same airspeed of 220 km/h.

Step-by-Step Solution

Answer

(a) Wind velocity is 44.72 km/h, 63.43° south of west. (b) Set course 12.53° north of west.

Step 1: Calculate Airplane's Ground Velocity

To find the ground velocity, we need to calculate the total displacement vector of the airplane. The westward displacement is 120 km, and the southward displacement is 20 km. The total displacement can be represented as the vector \( \mathbf{d} = (-120, -20) \) after 0.5 hours of flying.

Step 2: Determine Ground Velocity Components

The ground (actual) velocity of the airplane \( \mathbf{v}_{g} \) can be obtained by dividing the displacement vector by the time of travel. Thus, \( \mathbf{v}_{g} = \left(-\frac{120}{0.5}, -\frac{20}{0.5}\right) = (-240, -40) \) km/h. It means the actual velocity of the airplane in the west direction is 240 km/h and in the south direction is 40 km/h.

Step 3: Calculate Wind Velocity Components

The airplane's compass heading due west with an airspeed of 220 km/h means its air velocity components are \( \mathbf{v}_{a} = (-220, 0) \) km/h. Using the equation \( \mathbf{v}_{g} = \mathbf{v}_{a} + \mathbf{v}_{w} \), where \( \mathbf{v}_{w} \) is the wind velocity, the components of wind velocity can be found: \( \mathbf{v}_{w} = \mathbf{v}_{g} - \mathbf{v}_{a} = (-240, -40) - (-220, 0) = (-20, -40) \) km/h.

Step 4: Calculate Wind Velocity Magnitude and Direction

The magnitude of the wind velocity is \( \sqrt{(-20)^2 + (-40)^2} = \sqrt{400 + 1600} = \sqrt{2000} = 44.72 \) km/h. The direction relative to west can be found using the tangent function: \( \tan(\theta) = \frac{40}{20} = 2 \). Thus, \( \theta = \tan^{-1}(2) \approx 63.43^\circ \) south of west.

Step 5: Adjust Heading for Known Wind Velocity

For part (b), if the wind is 40 km/h due south, the pilot must adjust her heading to maintain a westward track. The components of her desired ground velocity are \( (-220, 0) \), given the plane travels directly west. Let \( \mathbf{v}_{h} = (x, y) \) be the air velocity needed to counteract the south wind. The total motion of the plane needs to cancel out the wind velocity's south component, so \( y = 40 \), and to achieve the west velocity of -220 km/h when wind is factored in, \( x + (-40) = -220 \) implies \( x = -180 \). The required heading is \( (-180, 40) \).

Step 6: Calculate the Correct Heading Magnitude and Direction

The heading magnitude is \( \sqrt{(-180)^2 + (40)^2} = \sqrt{32400 + 1600} = \sqrt{34000} = 184.39 \) km/h. To find the heading direction relative to west, use \( \theta = \tan^{-1}\left(\frac{40}{180}\right) = \tan^{-1}(0.2222) \approx 12.53^\circ \) north of west.

Key Concepts

Wind Velocity Calculation

Wind velocity refers to both the speed and direction of the wind acting on an object, in this case, an airplane. To determine the wind velocity, you start by establishing the plane's actual ground velocity after taking into account any deviations from its intended path. In this exercise, the wind has caused the plane to drift from its compass course directly west.
The total displacement of the plane was 120 km west and 20 km south in half an hour, meaning its ground velocity was -240 km/h in the westward direction and -40 km/h in the southward direction.

• **Ground Velocity Vector**: Calculated by dividing the displacement by time, turning the positional change into a velocity.
• **Air Velocity Vector**: Known from the compass heading, here simplified as \( \mathbf{v}_a = (-220, 0) \) km/h.
• **Wind Velocity Vector**: The difference between the ground and air velocities give us the wind velocity, \( \mathbf{v}_w = (-20, -40) \).

To find the wind's magnitude and direction, use Pythagorean Theorem and trigonometric functions. The magnitude is determined by \( \sqrt{(-20)^2 + (-40)^2} = 44.72 \) km/h. The direction is calculated using the arctan function \( \theta = \tan^{-1} \left(\frac{40}{20} \right) \approx 63.43^\circ \) south of west.

Ground and Airspeed Relationships

Understanding the relationship between ground speed and airspeed is crucial when navigating vehicles like airplanes. **Airspeed** is the speed of an aircraft relative to the air through which it moves, distinct from **ground speed**, which is the aircraft's speed over the ground.
In the problem, the pilot set an airspeed to maintain a fixed direction but found the airplane veered off course due to unknown winds. Original airspeed was 220 km/h toward due west, but wind changed the actual path.

• **Ground Speed**: For this scenario, calculated by the equation \( \mathbf{v}_g = \mathbf{v}_a + \mathbf{v}_w \), revealing a change due to wind velocity.
• **Adjustment for Wind**: Knowing wind effects is crucial to plan accurate headings, so pilots compute adjustments by reversing errors in ground velocity.

Without accommodating wind, the expected airspeed alone couldn't maintain a direct course due west. The exercise teaches us the significant influence wind has on navigation, specifically the correction methods needed to counteract these deviations. Pilots often moderate their compass heading to ensure the resultant ground track keeps them on course.

Trigonometry in Physics

Trigonometry is fundamental in physics for solving problems involving angles and distances, such as determining a plane's velocity and course adjustment. This exercise highlights how trigonometry assists in calculating both magnitude and directional angle of the wind affecting the airplane.
Using trigonometric functions, we can resolve vectors into components and calculate resultant vectors. This includes forces like wind or velocity, represented as directional vectors.

• **Magnitude Using Pythagorean Theorem**: Calculating overall effect from components using \( \sqrt{a^2 + b^2} \).
• **Direction via Inverse Tangent**: Finding angle direction by \( \theta = \tan^{-1}(\text{opposite over adjacent}) \).

For instance, after determining that southward and westward components are resultant wind vectors, trigonometry further reveals its angle relative to compass west, ensuring precision in subsequent heading corrections. Aviation relies heavily on such calculations, where any inaccuracies can result in significant deviations from the intended route. The practical application of trigonometry underscores how essential these math principles are in real-world physics problems.