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Chapter 1: Problem 89

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

Expert verified
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

01

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: \((62.5, 107.8)\)
02

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.
03

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.
04

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.

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