/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 The coordinates of a bird flying... [FREE SOLUTION] | 91Ó°ÊÓ

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The coordinates of a bird flying in the \(x y\) -plane are given by \(x(t)=\alpha t\) and \(y(t)=3.0 \mathrm{~m}-\beta t^{2},\) where \(\alpha=2.4 \mathrm{~m} / \mathrm{s}\) and \(\beta=1.2 \mathrm{~m} / \mathrm{s}^{2} .\) (a) Sketch the path of the bird between \(t=0\) and \(t=2.0 \mathrm{~s}\). (b) Calculate the velocity and acceleration vectors of the bird as functions of time. (c) Calculate the magnitude and direction of the bird's velocity and acceleration at \(t=2.0 \mathrm{~s}\). (d) Sketch the velocity and acceleration vectors at \(t=2.0 \mathrm{~s}\). At this instant, is the bird's speed increasing, decreasing, or not changing? Is the bird turning? If so, in what direction?

Short Answer

Expert verified
The bird's speed is decreasing since the acceleration points in the opposite direction to the motion. Because the velocity vector does not line up with the path tangent, the bird is turning left.

Step by step solution

01

Sketching the path

Plot the given \(x(t)\) and \(y(t)\) equations in a 2D plane for \(t = 0\) to \(2.0 s\).
02

Calculating Velocity and Acceleration vectors

The velocity vector components \(v_x\) and \(v_y\) can be found by differentiating the position equations. So, \(v_x(t) = \frac{dx}{dt} = \alpha\) and \(v_y(t) = \frac{dy}{dt} = -2\beta t\). For acceleration vector components \(a_x\) and \(a_y\), a second derivative is needed. So, \(a_x(t) = \frac{d^2x}{dt^2} = 0\) and \(a_y(t) = \frac{d^2y}{dt^2} = -2\beta\).
03

Calculate Magnitude and Direction of Velocity and Acceleration

At \(t=2.0s\), substitute \(t\) in the velocity and acceleration equations. The magnitude of these vectors can be computed as \(\sqrt{v_x^2 + v_y^2}\) and \(\sqrt{a_x^2 + a_y^2}\) respectively. The direction of these vectors can be calculated by finding the angle theta, \(\theta = tan^{-1}(\frac{v_y}{v_x})\) and \(\theta_a = tan^{-1}(\frac{a_y}{a_x})\) if necessary.
04

Sketching the vectors and describing the motion

Sketch the vectors for velocity and acceleration at time \(t = 2.0s\). From the acceleration vector, we can see that the acceleration is directed downward, implying that the bird is moving slower, known as decelerating. From the velocity vector and path, we can see that the bird is turning left.

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Key Concepts

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

Understanding Kinematics in Motion
Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause the motion. It describes the relationship between the positions, velocity, and acceleration of an object moving in two dimensions.

When analyzing the flight of a bird as in our exercise, the kinematic equations provide a method to calculate the bird's position \(x(t)\) and \(y(t)\) as functions of time \(t\). These equations help to generate the trajectory or path of the bird's flight. To sketch the bird's path, one plots the positions at various instants and connects these to visualize the motion.

The flight trajectory for our bird can be derived from the equations given: \(x(t)=\alpha t\) for horizontal motion and \(y(t)=3.0 \mathrm{~m}-\beta t^{2}\) for vertical motion, indicating a parabolic path - a key characteristic of projectile motion, which will be discussed in more detail in a later section.

This kinematic analysis allows students to visualize the bird's path and understand how position, velocity, and acceleration interrelate, which is an essential aspect of two-dimensional motion.
Velocity and Acceleration Vectors
To deepen our understanding of the bird's motion, we must consider the velocity and acceleration vectors. These vectors provide information not just about the speed, but also the direction of motion.

A velocity vector \(\vec{v}\) in two dimensions has two components: \(v_x\) and \(v_y\), representing horizontal and vertical motion, respectively. In our case, \(v_x(t) = \alpha\) remains constant since \(\alpha\) is a constant value, suggesting that the bird's horizontal speed doesn't change. Meanwhile, \(v_y(t) = -2\beta t\) changes with time, implying that the vertical speed does change.

The acceleration vector \(\vec{a}\) also comprises of two components: \(a_x\) and \(a_y\), with \(a_x\) being zero in our exercise because the horizontal motion is at constant velocity (no horizontal acceleration), and \(a_y = -2\beta\), pointing downward and indicating a constant vertical acceleration pointed towards the ground due to gravity.

Knowing how to calculate these vectors and their components is key to predicting and understanding the dynamic nature of the bird's flight, and is a core concept of motion in two dimensions. Students can use differentiation from calculus to find these values, as seen in the steps of the provided solution.
Decoding Projectile Motion
Projectile motion is a form of motion experienced by an object that is launched near the Earth's surface and moves along a curved path under the action of gravity only. The most important fact about projectile motion is that the horizontal and vertical motions are independent of each other.

In our exercise, at \(t=0\) to \(t=2.0 ~s\), the bird's flight represents a classic example of projectile motion. The equations given for \(x(t)\) and \(y(t)\) describe a parabolic trajectory, typical for projectiles. \(x(t)\) shows a linear motion due to the constant horizontal velocity, while \(y(t)\) shows a quadratic change due to the acceleration caused by gravity (here represented by \(\beta\)).

When calculating the bird's velocity and acceleration at any given time, one can determine not only the speed but also the direction of motion. For instance, the bird's turning motion is implicated by the direction of its acceleration vector and changing velocity vector components.

This concept reinforces the non-linear path that is characteristic of objects in projectile motion, which is essential for students to understand how motion in two dimensions can be analyzed and predicted. Understanding projectile motion allows students to apply these concepts to various real-world scenarios, such as sports, engineering, and even space travel.

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Most popular questions from this chapter

Firefighters use a high-pressure hose to shoot a stream of water at a burning building. The water has a speed of \(25.0 \mathrm{~m} / \mathrm{s}\) as it leaves the end of the hose and then exhibits projectile motion. The firefighters adjust the angle of elevation \(\alpha\) of the hose until the water takes \(3.00 \mathrm{~s}\) to reach a building \(45.0 \mathrm{~m}\) away. Ignore air resistance; assume that the end of the hose is at ground level. (a) Find \(\alpha\). (b) Find the speed and acceleration of the water at the highest point in its trajectory. (c) How high above the ground does the water strike the building, and how fast is it moving just before it hits the building?

The earth has a radius of \(6380 \mathrm{~km}\) and turns around once on its axis in \(24 \mathrm{~h}\). (a) What is the radial acceleration of an object at the earth's equator? Give your answer in \(\mathrm{m} / \mathrm{s}^{2}\) and as a fraction of \(g .\) (b) If \(a_{\mathrm{rad}}\) at the equator is greater than \(g\), objects will fly off the earth's surface and into space. (We'll see the reason for this in Chapter 5.) What would the period of the earth's rotation have to be for this to occur?

In a carnival booth, you can win a stuffed giraffe if you toss a quarter into a small dish. The dish is on a shelf above the point where the quarter leaves your hand and is a horizontal distance of \(2.1 \mathrm{~m}\) from this point (Fig. E3.19). If you toss the coin with a velocity of \(6.4 \mathrm{~m} / \mathrm{s}\) at an angle of \(60^{\circ}\) above the horizontal, the coin will land in the dish. Ignore air resistance. (a) What is the height of the shelf above the point where the quarter leaves your hand? (b) What is the vertical component of the velocity of the quarter just before it lands in the dish?

Two piers, \(A\) and \(B,\) are located on a river; \(B\) is \(1500 \mathrm{~m}\) down- stream from \(A\) (Fig. E3.38). Two friends must make round trips from pier \(A\) to pier \(B\) and return. One rows a boat at a constant speed of \(4.00 \mathrm{~km} / \mathrm{h}\) relative to the water; the other walks on the shore at a constant speed of \(4.00 \mathrm{~km} / \mathrm{h}\). The velocity of the river is \(2.80 \mathrm{~km} / \mathrm{h}\) in the direction from \(A\) to \(B\). How much time does it take each person to make the round trip?

A major leaguer hits a baseball so that it leaves the bat at a speed of \(30.0 \mathrm{~m} / \mathrm{s}\) and at an angle of \(36.9^{8}\) above the horizontal. Ignore air resistance. (a) At what \(t w o\) times is the baseball at a height of \(10.0 \mathrm{~m}\) above the point at which it left the bat? (b) Calculate the horizonta and vertical components of the baseball's velocity at each of the two times calculated in part (a). (c) What are the magnitude and directior of the baseball's velocity when it returns to the level at which it lef the bat?

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