/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Describes the position of a movi... [FREE SOLUTION] | 91Ó°ÊÓ

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Describes the position of a moving body at time \(t\). Determine whether, at time \(t=4,\) the body is moving forward, backward, or neither. $$ p(t)=t^{2}-8 t $$

Short Answer

Expert verified
At \( t=4 \), the body is neither moving forward nor backward, as the velocity is zero.

Step by step solution

01

- Find the Velocity Function

To determine the direction of movement, we first need to find the derivative of the position function, which is the velocity function. Given the position function is \( p(t) = t^2 - 8t \), we differentiate with respect to \( t \):\[p'(t) = \frac{d}{dt}(t^2 - 8t) = 2t - 8.\]
02

- Evaluate the Velocity at \( t = 4 \)

Once we have the velocity function, \( p'(t) = 2t - 8 \), substitute \( t = 4 \) to find the velocity at this specific time:\[p'(4) = 2(4) - 8 = 8 - 8 = 0.\]
03

- Determine Direction of Movement

Using the value of the velocity at \( t=4 \), evaluate whether the body is moving forward, backward, or neither. Since \( p'(4) = 0 \), the velocity is zero, indicating that the body is neither moving forward nor backward at \( t=4 \).

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Key Concepts

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

Understanding the Derivative of a Function
The derivative of a function is a core principle in calculus, especially when it comes to understanding motion. Derivatives allow us to find rates of change. When dealing with movement, the derivative of the position function gives us the velocity function. In our exercise, the position function is given as \( p(t) = t^2 - 8t \). To find how fast an object is moving, or its velocity, we differentiate the position function with respect to \( t \). This means calculating the derivative, which involves applying basic differentiation rules: the power rule for the \( t^2 \) term and the constant multiple rule for the \( 8t \) term. The derivative is computed as \( p'(t) = 2t - 8 \). This new function, \( p'(t) \), is what we use to determine the object's velocity at any point in time. By understanding derivatives, we can thus predict how the object's position changes, which is fundamental in analyzing motion.
Evaluating Functions to Determine Behavior
Evaluating a function is crucial when we need precise information about motion at a specific time. Once we have the velocity function \( p'(t) = 2t - 8 \), the next step is to determine the object's behavior at a particular moment, using this velocity function. In the exercise, we're interested in time \( t = 4 \). To evaluate, we substitute \( t \) with 4 in the velocity function to get \( p'(4) = 2(4) - 8 \). Calculations yield a result of \( 0 \), which tells us that at \( t=4 \), the velocity of the object is zero. Thus, evaluating velocity functions at specific times helps reveal whether the object is moving or at rest. This practical use of evaluating functions is especially helpful in deciding movement direction or identifying stationary points.
Interpreting Motion in Calculus
Interpreting motion in calculus relies on understanding the relationship between position, velocity, and acceleration. When we find the derivative of a position function, we shift our perspective from simply knowing where an object is, to understanding how it moves. The result from our evaluation \( p'(4) = 0 \) signifies that the object's velocity is zero at time \( t=4 \). In practical terms, this means the object is momentarily at rest. Exploring deeper into calculus, had the velocity been positive, it would indicate forward movement. Conversely, a negative velocity would depict backward motion. By examining these signs — positive, negative, or zero — we can sketch a complete picture of the object's motion over time. Thus, calculus not only pinpoints motion direction but also its magnitude and changes, enhancing our comprehension of dynamic systems.

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Most popular questions from this chapter

An initial value problem is given, along with its exact solution. (Read the instructions for Exercises \(47-50\) for terminology.) Verify that the given solution is correct by substituting it into the given differential equation and the initial value condition. Calculate the Euler's Method approximation \(y_{1}=y_{0}+F\left(x_{0}, y_{0}\right) \Delta x\) of \(y\left(x_{1}\right)\) where \(\Delta x=x_{1}-x_{0} .\) Let \(m_{1}=\left(F\left(x_{0}, y_{0}\right)+F\left(x_{1}, y_{1}\right)\right) / 2\) and \(z_{1}=y_{0}+\) \(m_{1} \Delta x .\) This is the Improved Euler Method approximation of \(y\left(x_{1}\right) .\) Calculate \(z_{1} .\) By evaluating \(y\left(x_{1}\right),\) determine which of the two approximations, \(y_{1}\) or \(z_{1},\) is more accurate. $$ \begin{array}{l} d y / d x=x+y, y(1)=2, x_{1}=1.2 ; \text { Exact solution: } y(x)= \\ 4 \exp (x-1)-x-1 \end{array} $$

An initial value problem is given, along with its exact solution. (Read the instructions for Exercises \(47-50\) for terminology.) Verify that the given solution is correct by substituting it into the given differential equation and the initial value condition. Calculate the Euler's Method approximation \(y_{1}=y_{0}+F\left(x_{0}, y_{0}\right) \Delta x\) of \(y\left(x_{1}\right)\) where \(\Delta x=x_{1}-x_{0} .\) Let \(m_{1}=\left(F\left(x_{0}, y_{0}\right)+F\left(x_{1}, y_{1}\right)\right) / 2\) and \(z_{1}=y_{0}+\) \(m_{1} \Delta x .\) This is the Improved Euler Method approximation of \(y\left(x_{1}\right) .\) Calculate \(z_{1} .\) By evaluating \(y\left(x_{1}\right),\) determine which of the two approximations, \(y_{1}\) or \(z_{1},\) is more accurate. $$ \begin{array}{l} d y / d x=x-y, y(-2)=-1, x_{1}=-2.15 ; \text { Exact solution: } \\ y(x)=2 \exp (-2-x)+x-1 \end{array} $$

A function \(f,\) a point \(c,\) an increment \(\Delta x,\) and a positive integer \(n\) are given. Use the method of increments to estimate \(f(c+\Delta x)\). Then let \(h=\Delta x\) / \(N\). Use the method of increments to obtain an estimate \(y_{1}\) of \(f(c+h) .\) Now, with \(c+h\) as the base point and \(y_{1}\) as the value of \(f(c+h),\) use the method of increments to obtain an estimate \(y_{2}\) of \(f(c+2 h)\). Continue this process until you obtain an estimate \(y_{N}\) of \(f(c+N \cdot h)=f(c+\Delta x) .\) We say that we have taken \(N\) steps to obtain the approximation. The number \(h\) is said to be the step size. Use a calculator or computer to evaluate \(f(c+\Delta x)\) directly. Compare the accuracy of the one step and \(N\) -step approximations. $$ f(x)=x^{1 / 3}, c=27, \Delta x=0.9, N=3 $$

Differentiate the given expression with respect to \(x\). $$ \tanh (\ln (x+2)) $$

Differentiate the given expression with respect to \(x\). $$ \operatorname{arccot}\left(1 / x^{2}\right) $$

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