91Ó°ÊÓ

The position of a particle between \(t=0\) and \(t=2.00 \mathrm{s}\) 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 \(t=2.00 \mathrm{s}\) is the particle instantaneously at rest? Does your numerical result agree with the \(v_{x}-t\) graph in part \((a) ?\) 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 \(t=2.00 \mathrm{s}\) is the velocity of the particle instantancously 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 \(t=2.00 \mathrm{s} ?(\mathrm{f})\) At what time(s) between \(t=0\) and \(t=2.00 \mathrm{s}\) is the particle speeding \(u p\) at the greatest rate? At what time(s) between \(t=0\) and \(t=2.00 \mathrm{s}\) 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

Find Velocity Function

The velocity function, \( v_x(t) \), is the derivative of the position function \( x(t) \). By differentiating \( x(t) = (3.00 \mathrm{m}/\mathrm{s}^3)t^3 - (10.0 \mathrm{m}/\mathrm{s}^2)t^2 + (9.00 \mathrm{m}/\mathrm{s})t \), we get:\[ v_x(t) = \frac{d}{dt}[3.00 t^3 - 10.0 t^2 + 9.00 t] = 9.00 t^2 - 20.0 t + 9.00. \]

02

Find Acceleration Function

The acceleration function, \( a_x(t) \), is the derivative of the velocity function \( v_x(t) \). By differentiating \( v_x(t) = 9.00 t^2 - 20.0 t + 9.00 \), we find:\[ a_x(t) = \frac{d}{dt}[9.00 t^2 - 20.0 t + 9.00] = 18.0 t - 20.0. \]

03

Plot Graphs for Part (a)

To draw the \( x-t \), \( v_x-t \), and \( a_x-t \) graphs, use the expressions:- \( x(t) = 3.00 t^3 - 10.0 t^2 + 9.00 t \)- \( v_x(t) = 9.00 t^2 - 20.0 t + 9.00 \)- \( a_x(t) = 18.0 t - 20.0 \).Evaluate these functions for several values of \( t \) between 0 and 2 and plot accordingly.

04

Find When Particle is at Rest (Part b)

The particle is at rest when \( v_x(t) = 0 \). Solve for \( t \) using: \[ 9.00 t^2 - 20.0 t + 9.00 = 0. \]Use the quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 9.00 \), \( b = -20.0 \), \( c = 9.00 \), to find the roots.

05

Calculate Acceleration at Rest Times (Part b)

Insert the \( t \) values from Step 4 into \( a_x(t) = 18.0 t - 20.0 \) to determine the acceleration. Confirm the type (positive or negative) of the acceleration matches the slope of \( v_x(t) \) at these points.

06

Determine When Velocity Isn't Changing (Part d)

The velocity is not changing when the acceleration \( a_x(t) = 0 \). Solve for \( t \) using: \[ 18.0 t - 20.0 = 0 \], which gives \( t = \frac{20.0}{18.0} = \frac{10}{9} \approx 1.11 \mathrm{s} \). Verify on the \( v_x-t \) and \( a_x-t \) graphs.

07

Find Greatest Distance from Origin (Part e)

To find the greatest distance from the origin, evaluate \( x(t) \) over the interval \( t = 0 \) to \( t = 2.00 \mathrm{s} \), and identify maximums through derivative testing. Solve \( v_x(t) = 0 \) to check critical points.

08

Find Maximum Speed Up and Slow Down Times (Part f)

The particle speeds up and slows down at the greatest rate when the magnitude of \( a_x(t) \) is at its maximum. As \( a_x(t) \) is linear, focus on the endpoints: calculate \( a_x(0) = -20.0 \) and \( a_x(2.00) = 16.0 \). Note these calculations on \( v_x-t \) and \( a_x-t \) graphs.

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Key Concepts

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

Position Function

In kinematics, the position function describes how the position of a particle changes over time. For the exercise at hand, the position function is given by:\[ x(t) = (3.00 \ \mathrm{m/s^3})t^3 - (10.0 \ \mathrm{m/s^2})t^2 + (9.00 \ \mathrm{m/s})t. \]This polynomial function of time shows the path of the particle as it moves. Each term represents a component of motion where:

• The first term \((3.00 \ \mathrm{m/s^3})t^3\) accounts for the motion resulting from acceleration (a cubic term).
• The second term \((10.0 \ \mathrm{m/s^2})t^2\) represents uniform acceleration.
• The third term \((9.00 \ \mathrm{m/s})t\) stands for initial velocity associated with linear motion.

To understand this function visually, one can plot \(x\) as a function of \(t\), showing how the position changes over the time interval from 0 to 2 seconds.

Velocity Function

The velocity function is derived by differentiating the position function with respect to time. It gives us the rate at which the particle's position changes. For the given problem, the velocity function is calculated as:\[ v_x(t) = \frac{d}{dt}\left[3.00t^3 - 10.0t^2 + 9.00t\right] = 9.00t^2 - 20.0t + 9.00. \]This quadratic function reflects how the velocity develops over time. Each of its terms gives insight into:

• The initial velocity of the particle, shown by the constant term \(9.00\), which is the velocity at \(t = 0\).
• The acceleration of the particle, represented by the \(-20.0t\) term, which connects to how velocity decreases over time.
• The rate of change of acceleration, given by the term \(9.00t^2\), indicating varying acceleration over time.

The velocity graph \(v_x-t\) provides a visual understanding of speed changes throughout the motion.

Acceleration Function

Acceleration is the derivative of the velocity function and describes how quickly the velocity of a particle changes. For this problem, the acceleration function is:\[ a_x(t) = \frac{d}{dt}[9.00t^2 - 20.0t + 9.00] = 18.0t - 20.0. \]This linear function tells us that acceleration changes uniformly over time. Here's what each part represents:

• The time-dependent term \(18.0t\) indicates a constant increase in acceleration as time progresses.
• The constant \(-20.0\) shows the initial acceleration effect when \(t = 0\).

The \(a_x-t\) graph will be a straight line, which aids in visually identifying when the particle speeds up or slows down in the interval from 0 to 2 seconds.

Graphing Kinematic Functions

To visualize kinematic concepts, graphing the position, velocity, and acceleration functions provides clarity about how a particle moves. Here's a brief breakdown:

• The \(x-t\) graph represents changes in position. It helps identify where the particle is located at any given time.
• The \(v_x-t\) graph identifies both the magnitude and signs of velocity. It reveals when the particle is moving forward, reversing, or remaining stationary.
• The \(a_x-t\) graph highlights how the particle accelerates or decelerates. It's particularly useful for recognizing when velocity changes are the greatest.

By plotting these functions over the same time interval, you better understand the particle's overall motion dynamics.

Instantaneous Rest

A particle is instantaneously at rest when its velocity is zero. This means the position function isn't changing at that moment in time, as per this exercise. To find when the particle is at rest, solve the equation:\[ 9.00t^2 - 20.0t + 9.00 = 0. \]Using the quadratic formula, you identify the exact times when \(v_x(t) = 0\). These points indicate where the tangent line on the \(x-t\) graph is horizontal.
Verify these instances on the velocity graph. Check if the acceleration \(a_x(t)\) at these points is consistent with the slope change sensed on the \(v_x-t\) graph to ensure complete accuracy in predictions.
Understanding instantaneous rest provides insight into the particle's transition and direction changes during its kinematic journey.