91Ó°ÊÓ

A turtle crawls along a straight line, which we will call the \(x\) -axis with the positive dircction to the right. The equation for the turtle's position as a function of time is \(x(t)=50.0 \mathrm{cm}+\) \((2.00 \mathrm{cm} / \mathrm{s}) t-\left(0.0625 \mathrm{cm} / \mathrm{s}^{2}\right) t^{2},(\mathrm{a})\) Find the turte's initial velocity, initial position, and initial acceleration. (b) At what time \(t\) is the velocity of the turtle zero? (c) How long after starting does it take the turtle to return to its starting point? (d) At what times \(t\) is the turtle a distance of 10.0 \(\mathrm{cm}\) from its starting point? What is the velocity (magnitude and direction) of the turtle at each of these times? (e) Sketch graphs of \(x\) versus \(t, v_{x}\) versus \(t,\) and \(a_{x}\) versus \(t\) for the time interval \(t=0\) to \(t=40 \mathrm{s}\)

Step by step solution

01

Identify Initial Position, Velocity, and Acceleration

The position equation is given as \( x(t) = 50.0 \text{ cm} + (2.00 \text{ cm/s}) t - (0.0625 \text{ cm/s}^2) t^2 \). Recognize that this is a quadratic equation of the form \( x(t) = x_0 + v_{x0} t + \frac{1}{2} a_x t^2 \).- Initial position \( x_0 = 50.0 \text{ cm} \).- Initial velocity \( v_{x0} = 2.00 \text{ cm/s} \).- Initial acceleration can be derived from the coefficient of \( t^2 \): \( \frac{1}{2} a_x = 0.0625 \text{ cm/s}^2 \). Therefore, \( a_x = -0.125 \text{ cm/s}^2 \).

02

Find Time When Velocity is Zero

The velocity \( v_x(t) \) can be found by differentiating \( x(t) \):\[ v_x(t) = \frac{d}{dt} (50.0 + 2.00 t - 0.0625 t^2) = 2.00 - 0.125 \times 2t = 2.00 - 0.125 \times 2t = 2.00 - 0.125 t \]To find when velocity is zero:\[ 0 = 2.00 - 0.125 t \]Solve for \( t \):\[ t = \frac{2.00}{0.125} = 16 \text{ s} \].

03

Calculate Time to Return to Starting Point

The turtle returns to the starting point when \( x(t) = x_0 \) or \( 50.0 \text{ cm} \):\[ 50.0 = 50.0 + 2.00 t - 0.0625 t^2 \]\( 0 = 2.00 t - 0.0625 t^2 \).Factor out \( t \) from the equation:\[ t (2.00 - 0.0625 t) = 0 \].Thus, the non-zero solution is:\[ t = \frac{2.00}{0.0625} = 32 \text{ s} \].

04

Find Times When Distance is 10 cm

The turtle is 10.0 cm away from starting position when \( |x(t) - 50.0| = 10.0 \):\[ x(t) = 40.0 \text{ cm} \text{ or } 60.0 \text{ cm} \].1. For \( x(t) = 40.0 \text{ cm} \):\[ 40.0 = 50.0 + 2.00 t - 0.0625 t^2 \]\[-10 = 2.00 t - 0.0625 t^2 \].Solve the quadratic equation:\( 0.0625 t^2 - 2.00 t - 10 = 0 \).Using quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 0.0625 \), \( b = -2.00 \), \( c = -10 \).This gives two solutions: \( t_1 = 5.3 \text{ s} \) and later utilizing similar methods for the \( x(t) = 60.0 \text{ cm} \) we find \( t_2 = 26.7 \text{ s} \).

05

Determine Velocities at Given Times

Calculate velocity at \( t_1 = 5.3 \text{ s} \):\[ v_x(5.3) = 2.00 - 0.125 \times 5.3 = 1.3375 \text{ cm/s} \]Direction is positive (right).Calculate velocity at \( t_2 = 26.7 \text{ s} \):\[ v_x(26.7) = 2.00 - 0.125 \times 26.7 = -1.3375 \text{ cm/s} \]Direction is negative (left).

06

Sketch Graphs

For the graphs:- Plot \( x(t) \) using the quadratic equation, observing the parabola opening downwards.- For \( v_x(t) = 2.00 - 0.125t \), plot a straight line with a negative slope starting at \( v_{x0} = 2.00 \text{ cm/s} \).- Plot \( a_x(t) = -0.125 \text{ cm/s}^2 \) as a constant line (since acceleration does not change with time in this case) for the time interval from \( t=0 \) to \( t=40 \text{ s} \).

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Key Concepts

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

Quadratic Motion

In kinematics, the term quadratic motion refers to a type of motion described by a quadratic equation, typically in the form of a position function that depends on time. In this case, the equation is given as \( x(t) = 50.0 \text{ cm} + (2.00 \text{ cm/s}) t - (0.0625 \text{ cm/s}^2) t^2 \).
This equation indicates that the turtle's movement is influenced by both linear and quadratic components.
Let's break this down:

• The initial position \( x_0 \) is 50.0 cm, indicating that this is the starting point on the x-axis from where the turtle begins its journey.
• The term \( 2.00 \text{ cm/s} \times t \) is linear, representing the initial velocity impacting the turtle's movement in a time-dependent manner.
• The \( -0.0625 \text{ cm/s}^2 \times t^2 \) term accounts for the effect of acceleration, causing the position to change in a quadratic manner as time progresses.

Detecting quadratic motion in an equation such as this one helps us anticipate key features of the turtle's path, such as its highest or lowest point, identified when the velocity reaches zero.

Velocity Calculation

Velocity calculation is crucial for understanding how fast and in which direction the turtle is moving at any given time. The velocity function is derived by taking the derivative of the position function.
In this example, the position function \( x(t) = 50.0 + 2.00t - 0.0625t^2 \) can be differentiated to find the velocity function \( v_x(t) \).
This results in:

• \( v_x(t) = \frac{d}{dt}(50.0 + 2.00t - 0.0625t^2) = 2.00 - 0.125t \).

The velocity function tells us how the speed and direction of the turtle change over time. In this specific case, we can determine that:

• The initial velocity \( v_{x0} = 2.00 \text{ cm/s} \), meaning the turtle starts moving to the right.
• The velocity decreases over time due to the negative acceleration, as indicated by the \( -0.125t \) term.

By analyzing \( v_x(t) \), we can also find when the turtle stops (velocity equals zero), by setting \( v_x(t) = 0 \) and solving for \( t \). This yields \( t = 16 \text{ s} \).
This information is valuable for knowing the time at which the turtle's speed is null.

Position Function

In kinematics, the position function reveals how an object's location changes over time. The given position function for the turtle is \( x(t) = 50.0 + 2.00t - 0.0625t^2 \). Understanding this helps assess the turtle's motion in detail.
The function is composed of:

• An initial position \( x_0 = 50.0 \text{ cm} \), which is the point of origin.
• A linear term \( (2.00 \text{ cm/s}) t \), indicating the constant velocity component contributing to a steady increase or decrease in position along the x-axis, hence reflecting the turtle's constant motion.
• A quadratic term \( - (0.0625 \text{ cm/s}^2) t^2 \), which counters the linear motion by introducing a decelerating effect due to negative acceleration.

The position function, therefore, is impactful in predicting when the turtle returns to its starting position. This happens when \( x(t) = x_0 \), yielding the key insight that at \( t = 32 \text{ s} \), the turtle returns."
Moreover, to find time-values when the turtle reaches certain points (like 10 cm away), the position function aids in solving for those instances by setting \( x(t) \) equal to the desired positions such as 40.0 cm and 60.0 cm.

Acceleration Analysis

Acceleration analysis in this context helps understand how the turtle's speed is affected over time. It is directly linked to the second derivative of the position function, but here we have directly identified it from the position equation.
The function reveals a constant acceleration, given as:

• From the term \( -0.0625 \text{ cm/s}^2 \), we identify that \( \frac{1}{2} a_x = 0.0625 \text{ cm/s}^2 \), logically deducing \( a_x = -0.125 \text{ cm/s}^2 \).
• This negative acceleration implies the turtle is decelerating as it moves along the x-axis.

Understanding this deceleration is vital because it indicates how quickly the velocity decreases over time. At any point, knowing the acceleration helps to anticipate subsequent velocity changes.
The constant nature of this acceleration is reflected in the straight-line plot of \( a_x(t) \) over time, emphasizing its unchanging rate regardless of the passage of time.