Question

A particle's motion is described by the following two parametric equations: $$ \begin{array}{l} x(t)=5 \cos (2 \pi t) \\ y(t)=5 \sin (2 \pi t) \end{array} $$ where the displacements are in meters and \(t\) is the time, in seconds. a) Draw a graph of the particle's trajectory (that is, a graph of \(y\) versus \(x\) ). b) Determine the equations that describe the \(x\) - and \(y\) -components of the velocity, \(v_{x}\) and \(v_{y}\), as functions of time. c) Draw a graph of the particle's speed as a function of time.

Short answer

Question: Given the parametric equations of a particle in motion as x(t) = 5cos(2πt) and y(t) = 5sin(2πt), draw the graph of the particle's trajectory, find the equations for the x and y components of its velocity as functions of time, and graph the particle's speed as a function of time. Answer: The particle's trajectory can be plotted using x(t) = 5cos(2πt) and y(t) = 5sin(2πt): The x and y components of the velocity as functions of time are: $$ v_x(t) = -10 \pi\sin(2\pi t) $$ $$ v_y(t) = 10 \pi\cos(2\pi t) $$ The particle's speed as a function of time is a constant value: $$ v(t) = 10\pi $$

Step-by-step solution

Plot Trajectory

To plot the particle's trajectory as described by its parametric equations for x(t) and y(t), use any graphing software or calculator and plug in the given equations. The graph should display x(t) = 5cos(2πt) on the x-axis and y(t) = 5sin(2πt) on the y-axis. #b) Determine the equations that describe the x- and y-components of the velocity#

Differentiate x(t) and y(t)

First, find the derivatives of x(t) and y(t) with respect to time (t) to obtain the x and y components of the velocity: $$ v_x(t)=\frac{dx(t)}{dt} \text{ and } v_y(t)=\frac{dy(t)}{dt} $$

Compute v_x(t) and v_y(t)

Differentiate x(t) and y(t) with respect to t: $$ v_x(t) = \frac{d}{dt}(5\cos(2\pi t))= -10\pi\sin(2\pi t) $$ $$ v_y(t) = \frac{d}{dt}(5\sin(2\pi t))= 10\pi\cos(2\pi t) $$ Thus, the x and y components of velocity, \(v_x\) and \(v_y\) as functions of time are: $$ v_x(t) = -10 \pi\sin(2\pi t) $$ $$ v_y(t) = 10 \pi\cos(2\pi t) $$ #c) Draw a graph of the particle's speed as a function of time#

Calculate speed

To get the particle's speed as a function of time, we need to find the magnitude of the velocity vector, which is the square root of the sum of the squares of the components, i.e., $$ v(t)=\sqrt{v_x(t)^2 + v_y(t)^2} $$

Compute v(t)

Plug in the expressions for \(v_x(t)\) and \(v_y(t)\) into the formula for \(v(t)\): $$ v(t) = \sqrt{(-10 \pi\sin(2\pi t))^2 + (10\pi\cos(2\pi t))^2} $$ $$ v(t) = \sqrt{100\pi^2(\sin^2(2\pi t) + \cos^2(2\pi t))} $$ Since \(\sin^2(2\pi t) + \cos^2(2\pi t)=1\), then the expression for the speed simplifies to: $$ v(t)=10\pi $$

Plot Speed

To plot the particle's speed as a function of time, plot \(v(t)=10\pi\) on the y-axis versus t on the x-axis. Notice that the speed is constant (doesn't change with time), as the graph will be a horizontal straight line at height 10π.

Key concepts

Understanding Particle Trajectory

In order to describe the path that a particle takes through space, we refer to its trajectory. Simply put, the trajectory is the line that maps out the journey of the particle from its starting point to its current or final position. In a two-dimensional space, this is often represented by a set of parametric equations, one for each spatial dimension (usually the x and y coordinates).

For instance, the exercise provides two parametric equations:

• \( x(t) = 5 \cos (2 \pi t) \)
• \( y(t) = 5 \sin (2 \pi t) \)

These equations, which vary with time (\( t \)), indicate that the particle is moving in a circular path with a radius of 5 meters. The cosine and sine functions dictate the points on the circle that the particle will be at any given time. When graphed with respect to each other for various values of \( t \), we obtain the trajectory of the particle, which in this case would be a perfect circle.

Breaking Down Velocity Components

Velocity, the rate of change of an object's position with respect to time, can also vary in different directions. That's where velocity components come into play. These are the separate velocities in each spatial dimension that collectively determine the object's overall velocity.

Say a particle is moving not just in a straight line, but in two dimensions—across the \( x \)-axis and \( y \)-axis. The \( x \)-component and \( y \)-component of the velocity (denoted as \( v_x \) and \( v_y \) respectively) are calculated by taking time derivatives of the position functions \( x(t) \) and \( y(t) \).

From our exercise:

• \( v_x(t) = -10 \pi \sin(2\pi t) \)
• \( v_y(t) = 10 \pi \cos(2\pi t) \)

These equations represent how fast the particle is moving in each respective direction at any given time, revealing a more nuanced understanding of its movement compared to just speed alone. The negative sign in \( v_x \) indicates direction, showing that the velocity in the \( x \)-direction will switch between positive and negative with the particle's rotation.

Interpreting a Time-Dependent Graph

Graphs are essential for visualizing how certain quantities change over time, a scenario often encountered when dealing with physics problems like particle motion. A time-dependent graph shows how some variable, such as speed, changes as time progresses.

In the provided exercise, we learn that the particle exhibits uniform circular motion, which is reflected in a flat, horizontal line when graphing speed as a function of time. The equation we use is: \( v(t) = 10\pi\). Since the speed value is constant and does not depend on \( t \), the graph will not rise or fall but remain at the constant value of \( 10\pi \) meters per second for any time \( t \). This visualization is powerful because it immediately lets students see that regardless of time, the particle's speed does not change, hence the motion is uniform, a fact not readily apparent without the use of the graph.