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Chapter 3: Problem 34

The nose of an ultralight plane is pointed due south, and its airspeed indicator shows 35 m/s. The plane is in a 10-m/s wind blowing toward the southwest relative to the earth. (a) In a vectoraddition diagram, show the relationship of \(\vec{v}_{P/E}\) (the velocity of the plane relative to the earth) to the two given vectors. (b) Let \(x\) be east and y be north, and find the components of \(\vec{v} _{P/E}\). (c) Find the magnitude and direction of \(\vec{v} _{P/E}\).

Short Answer

Expert verified
The plane's velocity relative to the earth is 42.67 m/s at 80.4掳 south of west.

Step by step solution

01

Understanding Velocity Vectors

First identify the given vectors. The velocity of the plane relative to air, \( \vec{v}_{P/A} \), is 35 m/s due south. The wind velocity, \( \vec{v}_{A/E} \), is 10 m/s toward the southwest. We need these vectors to find the velocity of the plane relative to the earth, \( \vec{v}_{P/E} \).
02

Setting up a Vector Addition Diagram

Draw \( \vec{v}_{P/A} \) as a vertical vector pointing due south. Next, draw \( \vec{v}_{A/E} \), which makes a 45掳 angle to the south-west direction. Use the head-to-tail method to add \( \vec{v}_{P/A} \) and \( \vec{v}_{A/E} \) for the resultant vector \( \vec{v}_{P/E} \).
03

Breaking Down Wind Vector into Components

Decompose \( \vec{v}_{A/E} \) (10 m/s) into its east-west and north-south components. Since it is directed southwest, \( v_{A/E,x} = -10 \cos(45^\circ) = -7.07 \) m/s (west) and \( v_{A/E,y} = -10 \sin(45^\circ) = -7.07 \) m/s (south).
04

Calculating the Components of Velocity Relative to Earth

The plane's velocity in the y-direction is the sum of its own velocity and the southward component of the wind: \( v_{P/E,y} = -35 + (-7.07) = -42.07 \) m/s. The plane's velocity in the x-direction comes entirely from the wind: \( v_{P/E,x} = -7.07 \) m/s.
05

Calculating the Magnitude of the Resultant Velocity

Use the Pythagorean theorem to calculate the magnitude: \(|\vec{v}_{P/E}| = \sqrt{(-7.07)^2 + (-42.07)^2} \approx 42.67\; \text{m/s}.\)
06

Calculating the Direction of the Resultant Velocity

Calculate the angle relative to the east: \( \theta = \tan^{-1}\left(\frac{-42.07}{-7.07}\right) \approx 80.4^\circ \) south of west.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Addition
In the world of physics, vector addition is a powerful tool to understand how different forces or directions combine. When the nose of a plane points due south with a speed of 35 m/s and the wind is blowing towards the southwest at 10 m/s, this is a classic case of vector addition in action. To solve the problem and find the plane's velocity relative to the earth (\( \vec{v}_{P/E} \)), we create a vector diagram. The technique often used is called the head-to-tail method. Here's how you do it:
  • Draw each vector starting with the first vector from a point. In this scenario, the plane鈥檚 velocity (\( \vec{v}_{P/A} \)) is a straight line pointing south.
  • Next, from the tip (or head) of this vector, draw the second vector, the wind velocity (\( \vec{v}_{A/E} \)), at a 45掳 angle southwest.
  • Finally, the resultant vector (\( \vec{v}_{P/E} \)) is drawn from the start of the first vector to the end of the last vector.
This head-to-tail method visually represents how differing directions and magnitudes of velocity combine to produce a total velocity.
Vector Components
Understanding the components of a vector is crucial when working with vectors that aren't aligned neatly along an axis. Vectors can be broken down into their perpendicular components鈥攗sually along the x (east-west) and y (north-south) axes.Let's break down the wind vector \( \vec{v}_{A/E} \), which is 10 m/s toward the southwest:
  • Because it makes a 45掳 angle with both the south and west directions, we use trigonometric functions to find its components:
  • The x-component, \( v_{A/E,x} \), is found by \(-10 \cos(45^\circ) = -7.07 \text{ m/s}\).
  • The y-component, \( v_{A/E,y} \), is \(-10 \sin(45^\circ) = -7.07 \text{ m/s}\).
Note that both components are negative, indicating the direction: west and south, respectively. Adding these components helps us understand how the overall motion is distributed along these axes, which is vital for session calculations.
Velocity Magnitude
Calculating the magnitude of a resultant velocity vector is a bit like finding the hypotenuse of a right triangle. After determining the x and y components of \( \vec{v}_{P/E} \), the magnitude is computed using the Pythagorean theorem:\[ |\vec{v}_{P/E}| = \sqrt{(-7.07)^2 + (-42.07)^2} \]Performing the calculations:
  • First, square each component: \((-7.07)^2\) and \((-42.07)^2\).
  • Add these squared values together.
  • Finally, take the square root of this sum to find the magnitude, \( |\vec{v}_{P/E}| \approx 42.67\; \text{m/s} \).
The magnitude tells us how fast the plane is actually moving over the ground, irrespective of direction.
Direction of Velocity
Understanding the direction of a velocity vector is critical in physics. Direction tells you where a vector points relative to a reference direction鈥攊n this case, relative to the earth鈥檚 axes.To find the direction of \( \vec{v}_{P/E} \):
  • The angle \( \theta \) is calculated using the inverse tangent function, \( \theta = \tan^{-1}\left(\frac{-42.07}{-7.07}\right) \).
  • Solving this gives \( \theta \approx 80.4^\circ \).
  • This angle is measured counter-clockwise, so the resultant direction is 80.4掳 south of west.
This angle indicates that the plane's path over the ground is primarily towards the south but slightly angled westwards as well. Understanding both magnitude and direction is crucial for a full grasp of how an object's motion translates in space.

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Most popular questions from this chapter

During a storm, a car traveling on a level horizontal road comes upon a bridge that has washed out. The driver must get to the other side, so he decides to try leaping the river with his car. The side of the road the car is on is 21.3 m above the river, while the opposite side is only 1.8 m above the river. The river itself is a raging torrent 48.0 m wide. (a) How fast should the car be traveling at the time it leaves the road in order just to clear the river and land safely on the opposite side? (b) What is the speed of the car just before it lands on the other side?

A "moving sidewalk" in an airport terminal moves at 1.0 m/s and is 35.0 m long. If a woman steps on at one end and walks at 1.5 m/s relative to the moving sidewalk, how much time does it take her to reach the opposite end if she walks (a) in the same direction the sidewalk is moving? (b) In the opposite direction?

A railroad flatcar is traveling to the right at a speed of 13.0 m/s relative to an observer standing on the ground. Someone is riding a motor scooter on the flatcar (\(\textbf{Fig. E3.30}\)). What is the velocity (magnitude and direction) of the scooter relative to the flatcar if the scooter's velocity relative to the observer on the ground is (a) 18.0 m/s to the right? (b) 3.0 m/s to the left? (c) zero?

A jet plane is flying at a constant altitude. At time \(t_1\) = 0, it has components of velocity \(v_x\) = 90 m/s, \(v_y\) = 110 m/s. At time \(t_2\) = 30.0 s, the components are \(v_x\) = -170 m/s, \(v_y\) = 40 m/s. (a) Sketch the velocity vectors at \(t_1\) and \(t_2\). How do these two vectors differ? For this time interval calculate (b) the components of the average acceleration, and (c) the magnitude and direction of the average acceleration.

A bird flies in the \(xy\)-plane with a velocity vector given by \(\overrightarrow{v}\) = \((a - \beta$$t^{2})\) \(\hat{l}+\gamma t \hat{j}\), with \(\alpha = 2.4 m/s\), \(\beta = 1.6 m/s^3\), and \(\gamma = 4.0 m/s^2\). The positive y-direction is vertically upward. At \(t = 0\) the bird is at the origin. (a) Calculate the position and acceleration vectors of the bird as functions of time. (b) What is the bird's altitude (y-coordinate) as it flies over \(x = 0\) for the first time after \(t = 0\)?
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