Question

A pilot decides to take his small plane for a Sunday afternoon excursion. He first flies north for 155.3 miles, then makes a \(90^{\circ}\) turn to his right and flies on a straight line for 62.5 miles, then makes another \(90^{\circ}\) turn to his right and flies 47.5 miles on a straight line. a) How far away from his home airport is he at this point? b) In which direction does he need to fly from this point on to make it home in a straight line? c) What was the farthest distance he was away from the home airport during the trip?

Short answer

b) What direction does the pilot need to fly from this point to make it home in a straight line? c) What is the farthest distance the pilot was away from the home airport during the trip? a) The pilot is approximately 123.97 miles away from his home airport after flying the prescribed path. b) The pilot needs to fly approximately 59.94 degrees from his current position to make it home in a straight line. c) The farthest distance the pilot was away from the home airport during the trip is approximately 166.19 miles.

Step-by-step solution

Find the position of the pilot after each movement

For each movement, we will add the displacement to his initial position. 1. Initial position: \((0,0)\) 2. After flying north for 155.3 miles: \((0, 155.3)\) 3. After making a \(90^{\circ}\) turn to the right and flying for 62.5 miles: \((62.5, 155.3)\) 4. After making another \(90^{\circ}\) turn to the right and flying for 47.5 miles: \((62.5, 107.8)\)

Calculate the distance from home airport

We can now calculate the distance from the home airport using the Pythagorean theorem: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) In this case, \((x_1, y_1) = (0,0)\) and \((x_2, y_2) = (62.5, 107.8)\). Therefore, $$d = \sqrt{(62.5 - 0)^2 + (107.8 - 0)^2} = \sqrt{62.5^2+107.8^2}$$ So, the distance: $$d \approx 123.97 \text{ miles}$$ a) The pilot is approximately 123.97 miles away from his home airport at this point.

Calculate the direction to fly home

To find the direction to fly home, we can use the tangent function to find the angle: $$\tan{\theta} = \frac{\text{opposite}}{\text{adjacent}} = \frac{107.8}{62.5}$$ Now, we calculate the angle in degrees: $$\theta = \arctan\left(\frac{107.8}{62.5}\right) \approx 59.94^{\circ}$$ b) The pilot needs to fly approximately \(59.94^{\circ}\) from his current position to make it home in a straight line.

Calculate the farthest distance from home

We now compare the distance from the starting point to the farthest point on the path he took. The farthest point is when he makes the first \(90^{\circ}\) turn and flies 62.5 miles. From the starting position \((0,0)\), this point is \((62.5,155.3)\). We calculate the distance between these points: $$d_{\text{farthest}} = \sqrt{(62.5 - 0)^2 + (155.3 - 0)^2} = \sqrt{62.5^2+155.3^2}$$ So, the farthest distance: $$d_{\text{farthest}} \approx 166.19 \text{ miles}$$ c) The farthest distance the pilot was away from the home airport during the trip is approximately 166.19 miles.

Key concepts

Pythagorean Theorem

The Pythagorean Theorem is a fundamental concept in mathematics, particularly useful for solving problems involving right triangles. It describes the relationship between the lengths of the sides of a right triangle. In a triangle with legs defined as \(a\) and \(b\), and hypotenuse defined as \(c\), the theorem states:

• \(c^2 = a^2 + b^2\)

To find the distance from the home airport in the exercise, this theorem helps us calculate the hypotenuse formed between the starting point and the pilot's final position. By plugging in the values of the displacements along the x-axis and y-axis into this equation, we can determine the straight-line distance the pilot is from the starting point. It beautifully simplifies the computation of an otherwise complex movement across different vectors into a straightforward calculation of lengths, illustrating why learning this theorem is invaluable.

Trigonometry

Trigonometry is the branch of mathematics dealing with the relationships between the angles and sides of triangles. In navigation or determining directions, trigonometry is often employed to find unknown angles and distances.

In our exercise, we use trigonometry to determine which direction the pilot needs to fly to return straight home. The tangent function, \(\tan(\theta)\), expresses the ratio of the opposite side to the adjacent side of a right triangle:

• \(\tan(\theta) = \frac{y}{x}\)

Here, the \(y\) component represents the vertical distance, and the \(x\) component represents the horizontal distance to home. By calculating the arc tangent, \(\arctan\), we derive the angle \(\theta\) indicating the direction the pilot must fly. Mastery of basic trigonometric functions such as sine, cosine, and tangent is crucial in many fields, including engineering, physics, and even computer graphics.

Displacement Calculation

Displacement is a vector quantity that refers to the change in position of an object. It has both magnitude and direction. In contrast to distance, which is scalar and only measures

the path length between two points, displacement provides a clearer picture of an object's overall movement. This distinction is key in physics and navigation.

In this example, to calculate the pilot's displacement from the home airport at any stage of the trip, we added up the vector components of each leg of the journey. The final displacement at the end of the flight is the distance from his final position to the starting point, computed using earlier

• distance formula derived from coordinates

Displacement gives us a concise idea of how far and in what direction the pilot is from the home.

Vector Components

Vector components are the projections of a vector along the axes of a coordinate system. They simplify handling two-dimensional motion by breaking vectors into the sum of individual one-dimensional motions.

• Each vector can be expressed in terms of i (x-direction) and j (y-direction) components.

When a plane moves in multiple directions, it experiences shifts in both x and y coordinates. By identifying these vector components, as done in the problem, we can easily compute combined movements.

Vector components allow us to restructure a problem from direct motion into horizontal and vertical directions, enabling calculations like using the Pythagorean theorem. Understanding vector components aids in solving complex real-world scenarios involving force, velocity, and displacements. It highlights the practicality of decomposing multi-directional problems into digestible bits, especially in physics and engineering disciplines.