/*! 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 coordinstes of a bird flying... [FREE SOLUTION] | 91Ó°ÊÓ

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The coordinstes 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 \(\beta=1.2 \mathrm{m} / \mathrm{s}^{2} \cdot\) (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 \(t=20 \mathrm{s}\) . (d) Sketch the velocity and acceleration vectors at \(t=2.0 \mathrm{s}\) . At this instant, is the bird speeding up, is it slowing down, or is its speed instantaneously not changing? Is the bird turning? If so, in what direction?

Short Answer

Expert verified
The bird is slowing down at \,\(t=2.0\)s, turning downwards.

Step by step solution

01

Calculate the Position Coordinates

The position coordinates of the bird as a function of time are given by the equations: \[ x(t) = \alpha t = 2.4t \text{ m}, \quad y(t) = 3.0 \text{ m} - \beta t^2 = 3.0 - 1.2t^2 \text{ m} \]We will use these equations to plot the coordinates over the interval \(t = 0\) to \(t = 2.0\) seconds.
02

Plot the Path of the Bird

Substitute the values of \(t\) from \(0\) to \(2.0\) seconds into the equations from Step 1 to get the position (x, y) at these times:- At \(t = 0\): \(x = 0\), \(y = 3.0\)- At \(t = 2.0\): \(x = 4.8\), \(y = -1.8\) Plotting these gives a parabolic path opening downwards.
03

Calculate Velocity Vectors

The velocity in the x-direction is the derivative of \(x(t)\):\[ v_x(t) = \frac{dx}{dt} = \alpha = 2.4 \text{ m/s} \]The velocity in the y-direction is the derivative of \(y(t)\):\[ v_y(t) = \frac{dy}{dt} = -2\beta t = -2(1.2)t = -2.4t \text{ m/s} \]
04

Calculate Acceleration Vectors

Acceleration is the derivative of velocity. Since \(v_x(t)\) is constant, the acceleration \(a_x(t)\) is:\[ a_x(t) = 0 \]For the y-component, the acceleration is:\[ a_y(t) = \frac{dv_y}{dt} = -2.4 \text{ m/s}^2 \]
05

Find Magnitude and Direction at t=2.0s

The velocity vector at \(t=2.0\)s has components \(v_x = 2.4\) m/s and \(v_y = -4.8\) m/s.- Magnitude of velocity: \[ |v| = \sqrt{(2.4)^2 + (-4.8)^2} = \sqrt{6.84} \approx 5.37 \text{ m/s} \]- Direction (angle \(\theta\)) using tangent: \[ \theta = \tan^{-1}\left(\frac{-4.8}{2.4}\right) = \tan^{-1}(-2) \approx -63.43^\circ \] Acceleration remains at \((0, -2.4)\) m/s² with magnitude 2.4 m/s² and direction straight downward.
06

Analyze Speed and Turning at t=2.0s

Since the velocity's magnitude is changing and the y-component is negative, the bird is slowing down in the y-direction. The bird is turning, as the path is parabolic and it's decelerating downwards. The downward acceleration indicates the bird is turning downwards.

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Key Concepts

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

Projectile Motion
Projectile motion refers to the movement of an object thrown or projected into the air, subject only to the acceleration of gravity. It beautifully combines both horizontal and vertical motions that occur independently of each other. The path an object follows, known as its trajectory, is typically parabolic.

The exercise we are analyzing revolves around a bird flying in a two-dimensional plane where its motion along the horizontal ( $x$) is linear due to a constant velocity, while its motion along the vertical ( $y$) involves a quadratic term. This signifies that while the bird's horizontal velocity remains constant, its vertical motion is affected by gravity, resembling the parabolic trajectory often seen in projectile motion.
Velocity and Acceleration Vectors
Understanding velocity and acceleration vectors is crucial for analyzing motion. A vector has both magnitude and direction, representing how velocity changes over time. It tells us not just how fast an object moves but also its moving direction.

In the given problem, the bird has a velocity component in the x-direction, \(v_x = 2.4 ext{ m/s}\), and in the y-direction, \(v_y = -2.4t ext{ m/s}\). Thus, the velocity vector is \( extbf{v}(t) = egin{pmatrix}2.4 \ -2.4t ext{ m/s} extbf{} ight)\). Acceleration, being the derivative of velocity, indicates changes in velocity. Since the bird moves with constant horizontal motion, its horizontal acceleration \(a_x\) is zero. However, the vertical acceleration is constant at \(a_y = -2.4 ext{ m/s}^2\), pointing downwards.
Kinematic Equations
Kinematic equations help us analyze motion by providing expressions to calculate unknown variables, given some initial parameters. They typically include terms for displacement, initial velocity, time, and acceleration. These equations are essential when studying objects in motion, especially when acceleration is constant, like gravity acting on a projectile.

In the bird's flight example, kinematic equations describe its position as $x(t) = 2.4t$ and $y(t) = 3.0 - 1.2t^2$. These depend on time, showing how the bird's position changes linearly over time in the horizontal direction and quadratically in the vertical direction, which results from the constant downward acceleration caused by gravity. Such insights are derived by taking derivatives for velocity and acceleration, making kinematic equations indispensable for solving complex motion problems.
Parabolic Trajectory
A parabolic trajectory is the path followed by an object under the influence of gravity, forming a parabolic shape. This occurs when the only force acting on an object is gravity, causing accelerated motion in one direction (typically vertical) while allowing uniform motion in another (horizontal).

For the bird in the exercise, its trajectory is parabolic due to its quadratic position in the vertical direction involving $y(t) = 3.0 - 1.2t^2$. The horizontal motion, unaffected by acceleration, maintains a straight line. By sketching the path between $t=0$ and $t=2.0$ seconds, we observe how the linear $x(t)$ combines with the quadratic $y(t)$, creating a bow that curves downward, characteristic of a parabolic flight. Such trajectories are not just theoretical; they model real-world phenomena like the arc of a basketball shot or a bird's coasting flight path.

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