Question

CALC The coordinates of a bird flying in the \(xy\)-plane are given by \(x(t)\) = \(at\) and \(y(t)\) = 3.0 m - \(\beta t^2\), where \(\alpha\) = 2.4 m/s and \(\beta\) = 1.2 m/s\(^2\). (a) Sketch the path of the bird between \(t\) = 0 and \(t\) = 2.0 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 s. (d) Sketch the velocity and acceleration vectors at \(t\) = 2.0 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

The bird's speed is increasing, and it is turning downward.

Step-by-step solution

Analyze the Equations for Position

Given the position equations as functions of time: \(x(t) = \alpha t\) and \(y(t) = 3.0 - \beta t^2\). Substitute \(\alpha = 2.4 \text{ m/s}\) and \(\beta = 1.2 \text{ m/s}^2\) to find the path. \(x(t) = 2.4t\) and \(y(t) = 3.0 - 1.2t^2\).

Sketch the Path

Plot the path of the bird for \(t\) ranging from 0 to 2.0 s. At \(t=0\), the position is \((0, 3.0)\). At \(t=2.0\), \(x = 4.8\) and \(y=-1.8\). Plot these points, creating a parabolic path with vertex at maximum height \((0, 3.0)\), moving rightwards and downwards.

Determine the Velocity Vectors

The velocity vector \(\mathbf{v}(t)\) is the derivative of the position vector. For \(x(t)\), \(v_x(t) = \frac{d}{dt}(2.4t) = 2.4 \text{ m/s}\). For \(y(t)\), \(v_y(t) = \frac{d}{dt}(3.0 - 1.2t^2) = -2.4t \text{ m/s}\). The velocity vector is \(\mathbf{v}(t) = (2.4, -2.4t)\).

Determine the Acceleration Vectors

The acceleration vector \(\mathbf{a}(t)\) is the derivative of the velocity vector. The derivative of \(v_x(t)\) is 0, so \(a_x(t) = 0\). The derivative of \(v_y(t)\) is \(-2.4 \text{ m/s}^2\), so \(a_y(t) = -2.4 \text{ m/s}^2\). The acceleration vector is \(\mathbf{a}(t) = (0, -2.4)\).

Magnitude and Direction at t=2.0 s

At \(t=2.0\), \(v_x(2) = 2.4 \text{ m/s}\) and \(v_y(2) = -4.8 \text{ m/s}\). Magnitude of velocity: \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{2.4^2 + (-4.8)^2} = 5.366 \text{ m/s}\). Direction: \(\theta = \tan^{-1}(\frac{-4.8}{2.4}) = -63.43^\circ\). For acceleration, \(a_x = 0\), \(a_y = -2.4\). Magnitude: \(a = 2.4 \text{ m/s}^2\) in the downward direction.

Sketch Velocity and Acceleration Vectors at t=2.0 s

For the velocity at \(t=2.0\), sketch a vector of length relative to 5.366 m/s directed downward at \(-63.43^\circ\) from the positive x-axis. Acceleration is a vector of length 2.4 m/s² downward.

Analyze Changes in Bird's Speed and Trajectory

The magnitude of the velocity increases from 5.0 m/s at \(t=0\) to 5.366 m/s at \(t=2.0\). Thus, speed is increasing. As the acceleration is in the negative y-direction, the bird is turning downwards due to the velocity component along with the downward acceleration.

Key concepts

Coordinate Systems

When studying the movement of objects, like the bird in our exercise, we often use the Cartesian coordinate system. This system involves two perpendicular axes, usually labeled as 'x' and 'y'. Here, the position of the bird over time is described using equations for both axes. The equation for the position in the x-axis is \(x(t) = 2.4t\), and for the y-axis, it is \(y(t) = 3.0 - 1.2t^2\). Each point on the path of the bird is identified by these coordinates at any given time.
This makes it easier to visualize and analyze the bird's motion, as it allows us to track its exact position on the plane at any given moment. Using coordinate systems in physics is essential, as it helps simplify complex motion into understandable data.

Velocity Vectors

The velocity vector provides important information about how fast the bird is moving and in what direction. It is derived by taking the derivative of the position equations for the bird. For the x-direction, the velocity is constant: \(v_x(t) = \frac{d}{dt}(2.4t) = 2.4\, \text{m/s}\). For the y-direction, the velocity changes with time: \(v_y(t) = \frac{d}{dt}(3.0 - 1.2t^2) = -2.4t\, \text{m/s}\).
A velocity vector is often represented as \((v_x, v_y) = (2.4, -2.4t)\). This shows the dynamics of motion—indicating that as time progresses, the y-component of the velocity becomes more negative, meaning the bird descends faster.

Acceleration Vectors

Acceleration vectors give insights into changes in the velocity of the bird. These are found by taking the derivative of the velocity vector components. In this problem, the x-component of acceleration is \(a_x(t) = 0\), showing no change in the horizontal velocity. Meanwhile, the y-component is \(a_y(t) = -2.4\, \text{m/s}^2\), meaning the bird is experiencing a consistent acceleration downwards.
Understanding acceleration helps predict future motion and determine if the motion of the bird is speeding up or changing direction. In this case, the downward acceleration suggests the bird is continuously being pulled down, indicative of gravitational influence.

Motion Analysis

Motion analysis involves examining various aspects of the bird's path, including its velocity and acceleration at different points in time. By calculating these vectors at specific times, such as \(t = 2.0\, \text{s}\), we get insights on speed direction and whether the bird's motion is dynamic or static.
At \(t = 2.0\, \text{s}\), the bird's velocity has a magnitude of \(5.366\, \text{m/s}\), with a direction pointing downwards at \(-63.43^{\circ}\). This motion indicates a downward trajectory, aligning with the negative y-velocity. The acceleration vector remains constant at \(2.4\, \text{m/s}^2\) downwards, showing a continuous pull in that direction.

Parabolic Trajectory

A parabolic trajectory is a common pathway in physics where an object follows a curved path due to the influence of forces like gravity. In our exercise, the bird's path is parabolic because of the quadratic nature of the y-coordinate equation, \(y(t) = 3.0 - 1.2t^2\).
The shape of this path results from the constant horizontal velocity and the acceleration acting solely in the vertical direction. As time progresses, the bird initially moves upward, reaches a peak, then descends—forming a characteristic parabola. This trajectory type is crucial for understanding projectile motion and can be observed in diverse phenomena, from simple projectiles to planets in space.