91Ó°ÊÓ

The position of a particle between \(t=0\) and \(t=2.00\) s is given by \(x(t)=\left(3.00 \mathrm{m} / \mathrm{s}^{3}\right) t^{3}-\left(10.0 \mathrm{m} / \mathrm{s}^{2}\right) t^{2}+\) \((9.00 \mathrm{m} / \mathrm{s}) t\) . (a) Draw the \(x-t, v_{x}-t,\) and \(a_{x}-t\) graphs of this particle. (b) At what time(s) between \(t=0\) and \(t=2.00\) s is the particle instantaneously at rest? Does your numerical result agree with the \(v_{x^{-}}\) graph in part (a)? (c) At each time calculated in part (b), is the acceleration of the particle positive or negative? Show that in each case the same answer is deduced from \(a_{x}(t)\) and from the \(v_{x}-t\) graph. (d) At what time(s) between \(t=0\) and \(t=2.00\) s is the velocity of the particle instantaneously not changing? Locate this point on the \(v_{x}-t\) and \(a_{x}-t\) graphs of part (a). (e) What is the particle's greatest distance from the origin \((x=0)\) between \(t=0\) and \(t=2.00 \mathrm{s} ?(\mathrm{f}) \mathrm{At}\) what time(s) between \(t=0\) and \(t=2.00\) s is the particle speeding \(u p\) at the greatest rate? At what time(s) between \(t=0\) and \(t=2.00\) s is the particle slowing down at the greatest rate? Locate these points on the \(v_{x}-t\) and \(a_{x}-t\) graphs of part (a).

Step by step solution

01

Determine the velocity function

To find the velocity as a function of time, differentiate the position function \(x(t)\) with respect to \(t\).\[ v_x(t) = \frac{dx}{dt} = \frac{d}{dt}\left((3.00 \text{ m/s}^3)t^3 - (10.0 \text{ m/s}^2)t^2 + (9.00 \text{ m/s})t\right) \]Using the power rule for differentiation, compute:\[ v_x(t) = 9.00 \text{ m/s}^3 \cdot t^2 - 20.0 \text{ m/s}^2 \cdot t + 9.00 \text{ m/s} \]

02

Determine the acceleration function

To find the acceleration as a function of time, differentiate the velocity function \(v_x(t)\) with respect to \(t\).\[ a_x(t) = \frac{dv_x}{dt} = \frac{d}{dt}\left(9.00 t^2 - 20.0 t + 9.00\right) \]Using the power rule again, compute:\[ a_x(t) = 18.0 t - 20.0 \]

03

Plot x-t, v-t, a-t graphs

Using a range of \(t\) values from \(0\) to \(2.00\) seconds:1. Create a table of values for \(x(t)\), \(v_x(t)\), and \(a_x(t)\) using the equations derived.2. Plot these values to create the \(x-t\), \(v_x-t\), and \(a_x-t\) graphs. This step involves computational/numerical methods using software or graphing tools.

04

Find time when particle is at rest

For the particle to be at rest, the velocity \(v_x(t)\) must be zero.Set the velocity equation to zero:\[ 9.00 t^2 - 20.0 t + 9.00 = 0 \]Use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 9.00\), \(b = -20.0\), \(c = 9.00\):\[ t = \frac{20.0 \pm \sqrt{(-20.0)^2 - 4 \times 9.00 \times 9.00}}{2 \times 9.00} \]Calculate the discriminant and solve for \(t\). The roots represent time(s) when the particle is at rest.

05

Analyze acceleration at rest time

Substitute the times found in Step 4 into the acceleration function \(a_x(t) = 18.0 t - 20.0\):Calculate \(a_x(t)\) for each root to determine the sign (positive or negative) of the acceleration.

06

Find time when velocity is not changing

For velocity to be not changing, acceleration must be zero:Set \(a_x(t)\) to zero:\[ 18.0 t - 20.0 = 0 \]Solve for \(t\):\[ t = \frac{20.0}{18.0} \approx 1.11 \text{ seconds} \]

07

Determine greatest distance from origin

Evaluate \(x(t)\) at the boundaries \(t = 0\) and \(t = 2.00\) seconds, and at the critical points (times when velocity \(v_x(t)=0\)) found in step 4.Compare these values to find the maximum \(x(t)\).

08

Find times when speeding up/slowing down is greatest

Speeding up or slowing down at the greatest rate implies maximum acceleration or deceleration. Consider the extreme points found from the graph of \(a_x(t)\).1. Find the maximum and minimum values of \(a_x(t)\) within the interval \(t = 0\) to \(t = 2.00\) seconds.2. Evaluate these at the critical points and endpoints.

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Key Concepts

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

Particle Motion

Understanding the motion of a particle involves exploring how it moves through space and time. In this exercise, the particle's position is described by a function of time, specifically a polynomial function. This means the equation gives us the particle's exact location on a one-dimensional path at any time between 0 and 2 seconds. Particle motion is a critical concept in kinematics, where we study how bodies move without considering the forces that cause the motion.

To analyze particle motion, we break it down into measurable parameters such as position, velocity, and acceleration. Each of these parameters provides insight into how the particle travels across its path. Observing these changes helps us to understand the dynamics of motion and also predict how the particle will behave at different points in time.

Velocity Function

The velocity function of a particle shows how fast and in what direction the particle is moving at any given time. In this problem, we derive the velocity function by differentiating the position function with respect to time. This is because velocity is the time rate of change of position, mathematically expressed as the first derivative of the position function. For the given function, the derived velocity equation is:

• \(v_x(t) = 9.00 \, ext{m/s}^3 \, t^2 - 20.0 \, ext{m/s}^2 \, t + 9.00 \, ext{m/s}\)

This quadratic equation helps us determine the velocity at any point in time between 0 and 2 seconds.

By setting this velocity equation to zero, we can also find when the particle is momentarily at rest. Solving for these specific times gives insights about the particle’s state of motion and plays a critical role in analyzing the overall motion using graphs.

Acceleration Function

Acceleration gives us an idea of how quickly the velocity of the particle is changing. To find the acceleration function from the velocity function, we take the derivative of the velocity function with respect to time. Mathematically, acceleration is the second derivative of the position function or the first derivative of the velocity function. For this exercise, the acceleration equation is:

• \(a_x(t) = 18.0 \, t - 20.0\)

Acceleration can be positive, negative, or zero, each indicating different aspects of motion. Positive acceleration means increasing speed, negative indicates decreasing speed, and zero means uniform motion.

Given its importance, checking the sign of the acceleration function at points where the particle is at rest will reveal whether the particle is starting to speed up or slow down as it changes direction. This aids in understanding the cyclic nature of motion.

Position Function

The position function describes the location of a particle at any given time. In our exercise, the position function is a cubic polynomial:

• \(x(t) = 3.00 \, ext{m/s}^3 \, t^3 - 10.0 \, ext{m/s}^2 \, t^2 + 9.00 \, ext{m/s} \, t\)

Using this function, we can determine where the particle is at any time between 0 and 2 seconds.

To find the particle's greatest distance from the origin, we examine the position function at the boundaries of the evaluated time range and also at critical points derived from where the velocity function equals zero. Comparing these values tells us the point at which the particle reaches its maximum displacement from the origin.

Graphing Motion

Graphing is an essential tool for visualizing the motion of the particle. It provides a clear representation of how position, velocity, and acceleration change over time. For this exercise, we create three key graphs:

• The Position-Time (x-t) graph shows where the particle is located at different times.
• The Velocity-Time (v-t) graph illustrates how fast the particle is moving and in which direction.
• The Acceleration-Time (a-t) graph displays how the rate of change of velocity varies.

Using these graphs, we can intuitively identify points where the particle changes direction (where velocity is zero), speeds up, and slows down.

Moreover, graph analysis helps verify our analytical findings, ensuring consistency between graphical and mathematical solutions.

Differentiation in Physics

Differentiation is a mathematical process that plays a vital role in physics, particularly in kinematics. It helps us find how quantities change over time, which is central to understanding motion. In this exercise, differentiation allows us to derive the velocity and acceleration functions from the position function.

Here’s how it works:

• The first derivative of the position function with respect to time gives the velocity function, explaining how position changes.
• The second derivative, or the first derivative of velocity, yields the acceleration function, providing insights into velocity changes.

This process simplifies analyzing complex motions by reducing motion description from position to velocity and finally to acceleration. Differentiation is thus crucial for predicting and interpreting the behavior of moving particles.