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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 speeding up, is it slowing down, or is its speed instantaneously not changing? Is the bird turning? If so, in what direction?

Step by step solution

01

Bird's Path Equation

The bird's path is defined by the parametric equations \( x(t) = \alpha t \) and \( y(t) = 3 - \beta t^2 \). Given that \( \alpha = 2.4 \text{ m/s} \) and \( \beta = 1.2 \text{ m/s}^2 \), these become \( x(t) = 2.4t \) and \( y(t) = 3 - 1.2t^2 \).

02

Sketch Bird's Path

Calculate important points: \( t = 0 \), \( x(0) = 0 \), \( y(0) = 3 \); \( t = 2 \), \( x(2) = 4.8 \), \( y(2) = 0.4 \). Connect these points and note the path is a downward curve.

03

Velocity Vector Calculation

Find velocity as the derivative of position. The velocity vector \( \vec{v}(t) = \left(\frac{dx}{dt}, \frac{dy}{dt}\right) = (2.4, -2.4t) \).

04

Acceleration Vector Calculation

Acceleration is the derivative of velocity. The acceleration vector \( \vec{a}(t) = \left(\frac{d^2x}{dt^2}, \frac{d^2y}{dt^2}\right) = (0, -2.4) \).

05

Velocity and Acceleration at t=2.0 s

Substitute \( t = 2 \) into velocity and acceleration: \( \vec{v}(2) = (2.4, -4.8) \), \( \vec{a}(2) = (0, -2.4) \).

06

Magnitude and Direction of Vectors

Velocity magnitude: \( |\vec{v}(2)| = \sqrt{2.4^2 + (-4.8)^2} = 5.37 \text{ m/s} \). Acceleration magnitude: \( |\vec{a}(2)| = |-2.4| = 2.4 \text{ m/s}^2 \). Velocity direction: \( \theta_v = \tan^{-1}\left(\frac{-4.8}{2.4}\right) = -63.4^\circ \). Acceleration direction is \(-90^\circ \).

07

Sketch Vectors at t=2.0 s

Draw velocity and acceleration vectors from a point. Velocity has components 2.4 m/s rightward and 4.8 m/s downward; acceleration is 2.4 m/s^2 downward. This shows downward movement.

08

Speed and Turning Analysis

Since both velocity and acceleration have downward components, the bird is speeding up in the downward direction. There is no change in horizontal speed (\( x \)-direction), so it turns downward.

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Key Concepts

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

Velocity Vector

In kinematics, the velocity vector describes the rate of change of the position of an object with respect to time. For the bird in our problem, the velocity vector \( \vec{v}(t) \) can be computed by taking the first derivative of the bird's position described by the parametric equations \( x(t) = 2.4t \) and \( y(t) = 3 - 1.2t^2 \). When we differentiate, we obtain \( \vec{v}(t) = \left(\frac{dx}{dt}, \frac{dy}{dt}\right) = (2.4, -2.4t) \).

The velocity vector elements tell us that:

• The horizontal component is a constant \(2.4 \text{ m/s}\), indicating constant rightward movement.
• The vertical component is \(-2.4t \), showing that the vertical speed increases in magnitude over time in the downward direction.

Understanding these components is crucial because they reveal how the bird's movement evolves over time in each direction.

Acceleration Vector

The acceleration vector is an essential concept in kinematics, representing the rate of change of velocity over time. This vector tells us how an object's speed and direction change. For the bird, we can calculate the acceleration vector \( \vec{a}(t) \) by deriving the velocity equations: \( \vec{a}(t) = \left( \frac{d^2x}{dt^2}, \frac{d^2y}{dt^2} \right) = (0, -2.4) \).

Key points about the acceleration vector include:

• The horizontal component is zero, indicating no change in the bird's horizontal velocity.
• The vertical component is \(-2.4 \text{ m/s}^2\), indicating a constant downward acceleration.

This downward acceleration affects how quickly the bird’s vertical velocity increases in the downward direction, offering insights into the impact of gravitational-like forces on its flight.

Parametric Equations

Parametric equations are a powerful tool in kinematics, allowing us to express an object's position in terms of a third parameter, usually time. In the bird's scenario, the parametric equations \( x(t) = 2.4t \) and \( y(t) = 3 - 1.2t^2 \) describe its motion in the plane.

Analyzing these equations, we can deduce:

• The parameter \( t \) represents time and provides a dynamic view of the bird's path.
• The function \( x(t) \) implies steady linear motion parallel to the x-axis.
• The function \( y(t) = 3 - 1.2t^2 \) suggests a quadratic change, indicating the path has a parabolic curve.

Using parametric equations offers a comprehensive view of the bird's movement, illustrating both the linear and quadratic influences on its path.

Path of Motion

The path of motion in kinematics describes the route or trajectory that an object follows as it moves through space. For the bird, analyzing its path has been made simpler due to its parametric equations \( x(t) = 2.4t \) and \( y(t) = 3 - 1.2t^2 \). This path can be visualized as a downward arc or curve.

To characterize this path:

• The points calculated at different time intervals, such as \( t=0 \) and \( t=2 \), give specific locations: (0, 3) and (4.8, 0.4), respectively.
• The trajectory connects these points, reflecting the constant horizontal shift and altering vertical descent.
• By graphing or sketching, you can see the bird starts at a height and subsequently descends in a curved path.

Understanding the path of motion helps predict future positions and analyze changes in velocity or acceleration over time, providing a holistic view of movement dynamics.