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Chapter 3: Problem 26

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

Expert verified
In this exercise, we looked at the motion of a particle described by the parametric equations \(x(t) = 5\cos(2\pi t)\) and \(y(t) = 5\sin(2\pi t)\). By examining the particle's trajectory, we found that it follows a circular path with a radius of 5 meters and period of 1 second. We differentiated the parametric equations to determine the x- and y-components of velocity, which are given by \(v_x(t) = -10\pi\sin(2\pi t)\) and \(v_y(t) = 10\pi\cos(2\pi t)\). Finally, we calculated the particle's speed, which is constant at 10π meters per second.

Step by step solution

01

Convert Parametric Equations to Position Vector

Combine the given parametric equations by considering them as a position vector \(\textbf{r}(t)\): $$ \textbf{r}(t) = x(t)\textbf{i} + y(t)\textbf{j} = 5\cos(2\pi t)\textbf{i} + 5\sin(2\pi t)\textbf{j}. $$
02

Eliminate Parameter t

Divide the y-coordinate equation by the x-coordinate equation to eliminate the parameter t: $$ \frac{y(t)}{x(t)} = \frac{5\sin(2\pi t)}{5\cos(2\pi t)} = \tan(2\pi t). $$ From the above equation, we have: $$ 2\pi t = \arctan\left(\frac{y(t)}{x(t)}\right). $$
03

Draw the Graph

From the given parametric equations, we have the following relations: $$ \begin{cases} x(t) = 5\cos(2\pi t)\\ y(t) = 5\sin(2\pi t) \end{cases}. $$ This represents a circle with the radius 5 meters and a period of 1 second. Plot the trajectory on the xy-plane by eliminating t using the above equation. #b) Equations for x- and y-components of Velocity#
04

Differentiate Parametric Equations

To find the x- and y-components of the velocity, differentiate the given parametric equations with respect to time t: $$ v_x(t) = \frac{dx(t)}{dt} = -10\pi\sin(2\pi t) $$ and $$ v_y(t) = \frac{dy(t)}{dt} = 10\pi\cos(2\pi t). $$ These equations describe the x- and y-components of the velocity as functions of time. #c) Graph of Particle's Speed as a Function of Time#
05

Calculate the Speed

The speed (magnitude of the velocity vector) can be calculated as: $$ v(t) = \sqrt{v_x(t)^2 + v_y(t)^2} = \sqrt{(-10\pi\sin(2\pi t))^2 + (10\pi\cos(2\pi t))^2}. $$ After simplifying the expression, we get: $$ v(t) = 10\pi. $$
06

Draw the Graph

From the above calculation, we can see that the speed is constant over time and is independent of the time t. Thus, the graph of the particle's speed as a function of time will be a horizontal line at the value \(v(t) = 10\pi\) meters per second.

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Most popular questions from this chapter

You passed the salt and pepper shakers to your friend at the other end of a table of height \(0.85 \mathrm{~m}\) by sliding them across the table. They both missed your friend and slid off the table, with velocities of \(5 \mathrm{~m} / \mathrm{s}\) and \(2.5 \mathrm{~m} / \mathrm{s}\), respectively. a) Compare the times it takes the shakers to hit the floor. b) Compare the distance that each shaker travels from the edge of the table to the point it hits the floor.

A cruise ship moves southward in still water at a speed of \(20.0 \mathrm{~km} / \mathrm{h},\) while a passenger on the deck of the ship walks toward the east at a speed of \(5.0 \mathrm{~km} / \mathrm{h}\). The passenger's velocity with respect to Earth is a) \(20.6 \mathrm{~km} / \mathrm{h}\), at an angle of \(14.04^{\circ}\) east of south. b) \(20.6 \mathrm{~km} / \mathrm{h}\), at an angle of \(14.04^{\circ}\) south of east. c) \(25.0 \mathrm{~km} / \mathrm{h}\), south. d) \(25.0 \mathrm{~km} / \mathrm{h}\), east. e) \(20.6 \mathrm{~km} / \mathrm{h}\), south.

A rock is tossed off the top of a cliff of height \(34.9 \mathrm{~m}\) Its initial speed is \(29.3 \mathrm{~m} / \mathrm{s}\), and the launch angle is \(29.9^{\circ}\) with respect to the horizontal. What is the speed with which the rock hits the ground at the bottom of the cliff?

A circus juggler performs an act with balls that he tosses with his right hand and catches with his left hand. Each ball is launched at an angle of \(75^{\circ}\) and reaches a maximum height of \(90 \mathrm{~cm}\) above the launching height. If it takes the juggler \(0.2 \mathrm{~s}\) to catch a ball with his left hand, pass it to his right hand and toss it back into the air, what is the maximum number of balls he can juggle?

A football is punted with an initial velocity of \(27.5 \mathrm{~m} / \mathrm{s}\) and an initial angle of \(56.7^{\circ} .\) What is its hang time (the time until it hits the ground again)?
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