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Chapter 3: Problem 83

A pilot starting from Athens, New York, wishes to fly to Sparta, New York, which is \(320 \mathrm{km}\) from Athens in the direction $20.0^{\circ} \mathrm{N}\( of \)\mathrm{E} .$ The pilot heads directly for Sparta and flies at an airspeed of \(160 \mathrm{km} / \mathrm{h}\). After flying for $2.0 \mathrm{h}$, the pilot expects to be at Sparta, but instead he finds himself \(20 \mathrm{km}\) due west of Sparta. He has forgotten to correct for the wind. (a) What is the velocity of the plane relative to the air? (b) Find the velocity (magnitude and direction) of the plane relative to the ground. (c) Find the wind speed and direction.

Short Answer

Expert verified
Answer: The velocity of the plane relative to the air is 160 km/h at an angle of 15.3掳 N of E. The velocity of the plane relative to the ground is 150 km/h at an angle of 21.4掳 N of E. The wind speed is 19.4 km/h, and the wind direction is west.

Step by step solution

01

Calculate the plane's intended displacement vector

The pilot wants to reach Sparta, which is 320 km away in the direction 20.0掳 N of E. To represent the plane's intended displacement, create a vector pointing 20.0掳 N of E with magnitude 320 km. Let the x-direction be east and the y-direction be north. Then, the components of the displacement vector are: x-component: \(V_{x} = 320 \cdot cos(20.0^\circ) = 301\text{ km}\) y-component: \(V_{y} = 320 \cdot sin(20.0^\circ) = 109\text{ km}\)
02

Calculate the actual displacement vector

Since the pilot is 20 km due west of Sparta after flying 2 hours, the actual displacement vector needs to be calculated. The x-component should be 20 km less than the intended x-component, and the y-component should be the same. Actual x-component: \(a_x = V_{x} - 20 = 301 - 20 = 281\text{ km}\) Actual y-component: \(a_y = V_{y} = 109\text{ km}\)
03

Calculate velocity relative to the ground

The velocity relative to the ground can be found by dividing the actual displacement by the time taken to cover that distance (2 hours). Velocity in x-direction: \(V_{gx} = \frac{a_x}{2} = \frac{281}{2} = 140.5\text{ km/h}\) Velocity in y-direction: \(V_{gy} = \frac{a_y}{2} = \frac{109}{2} = 54.5\text{ km/h}\)
04

Find the magnitude and direction

Now, we can find the magnitude and direction of the velocity vector relative to the ground. Magnitude: \(|V_g| = \sqrt{V_{gx}^2 + V_{gy}^2} = \sqrt{(140.5)^2 + (54.5)^2} \approx 150\text{ km/h}\) Direction: \(\theta = \arctan(\frac{V_{gy}}{V_{gx}}) = \arctan(\frac{54.5}{140.5}) \approx 21.4^\circ\text{ N}\text{ of E}\)
05

Calculate the velocity relative to the air

The plane's airspeed is given as 160 km/h. Now we need to find the direction of this velocity. We can use the law of cosines to find the angle \(\alpha\) between the intended and actual velocity vectors. \(\alpha = \arccos(\frac{V_{x}^2 + (V_{x}-20)^2 + V_{y}^2 + V_{y}^2 - 20^2}{2\cdot320^2}) \approx 4.7^\circ\) So, the velocity of the plane relative to the air is \(160\text{ km/h}\) at an angle of 20.0掳 - 4.7掳 = 15.3掳 N of E.
06

Calculate the wind velocity vector

Finally, we can calculate the wind velocity vector by finding the vector difference between the velocities relative to the air and relative to the ground. Wind velocity in x-direction: \(w_x = V_{gx} - 160 \cdot cos(15.3^\circ) \approx -19.4\text{ km/h}\) Wind velocity in y-direction: \(w_y = V_{gy} - 160 \cdot sin(15.3^\circ) \approx 0.0\text{ km/h}\)
07

Find the wind speed and direction

Wind speed: \(|w| = \sqrt{w_x^2 + w_y^2} = \sqrt{(-19.4)^2 + (0.0)^2} \approx 19.4\text{ km/h}\) Since \(w_y \approx 0\), the wind direction is along the negative x-direction, which is west. #Results#: (a) The velocity of the plane relative to the air is \(160\text{ km/h}\) at an angle of \(15.3^\circ\text{ N}\text{ of E}\). (b) The velocity of the plane relative to the ground is \(150\text{ km/h}\) at an angle of \(21.4^\circ\text{ N}\text{ of E}\). (c) The wind speed is \(19.4\text{ km/h}\), and the wind direction is west.

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