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Expert-verified Found in: Page 167 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Every morning Linda takes a thirty-minute jog in Central Park. Suppose her distance s in feet from the oak tree on the north side of the park $t$ minutes after she begins her jog is given by the function $s\left(t\right)$ shown that follows at the left, and suppose she jogs on a straight path leading into the park from the oak tree. (a) What was the average rate of change of Linda’s distance from the oak tree over the entire thirty-minute jog? What does this mean in real-world terms? (b) On which ten-minute interval was the average rate of change of Linda’s distance from the oak tree the greatest: the first $10$ minutes, the second $10$ minutes, or the last$10$ minutes?(c) Use the graph of $s\left(t\right)$ to estimate Linda’s average velocity during the $5$-minute interval from$t=5tot=10$. What does the sign of this average velocity tell you in real-world terms?(d) Approximate the times at which Linda’s (instantaneous) velocity was equal to zero. What is the physical significance of these times? (e) Approximate the time intervals during Linda’s jog that her (instantaneous) velocity was negative. What does a negative velocity mean in terms of this physical example? Ans:

(a) Therefore, the average rate of change of the person L of distance over an interval $\left[0,30\right]$ is $0$ . The average rate of change of distance over an interval [0,30] implies that the distance from a tree at $t=30$ is same as that of at $t=0$.

(b) Therefore, the average rate of change of distance over an interval $\left[20,30\right]$ is $-25$.

The average rate of change of distance is maximum on the first 10 minutes interval is $15$ feet per second.

(c) Average velocity is: $-26$

The negative value of average velocity indicates that the person moves towards the tree.

(d) It is observed that the slope of the function is zero at $t=4,t=12$ and $t=25$.

Thus, an approximated value of time t is $t=4$ minutes, $t=12$ minutes and $t=25$ minutes at which the velocity is zero. It implies that person L changes the direction of the position.

(e) The person L has negative velocity when the distance decreases.

The negative value of velocity indicates that person L moves towards the tree.

See the step by step solution

## Step 1. Given information.

Suppose her distance s in feet from the oak tree on the north side of the park $t$ minutes after she begins her jog is given by the function $s\left(t\right)$ shown that follows at the left and suppose she jogs on a straight path leading into the park from the oak tree.

## Step 2.  (a)  The main objective is to determine the average rate of change of the distance from the tree over the entire 30 minutes of jogging. Explain the meaning of the average rate of change of distance in the real world.

The average rate of change of a function $f\left(x\right)$ over a time interval $\left[a,b\right]$ is calculated as $\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

From the given graph, it is observed that the value of the distance at $t=0$ and $t=30$ is $0$

Use the formula localid="1648397091064" $\phantom{\rule{0ex}{0ex}}\frac{f\left(b\right)-f\left(a\right)}{b-a}$ to calculate the average rate of change of distance over an interval $\left[0,30\right]$.

$\begin{array}{r}\frac{f\left(30\right)-f\left(0\right)}{30-0}=\frac{0-0}{30}=0\\ \end{array}$

Therefore, the average rate of change of the person L of distance over an interval $\left[0,30\right]$ is $0$ . The average rate of change of distance over an interval $\left[0,30\right]$ implies that the distance from a tree at $t=30$ is the same as that of at $t=0$.

## Step 3. (b)  The main objective is to determine a 10 minutes interval on which the rate of change of distance is the greatest.

Consider the first $10$ minutes interval.

The distance from tree at t=0 is 0 and at $t=10$ is $150$ .

Use the formula localid="1648396781714" $\phantom{\rule{0ex}{0ex}}\frac{f\left(b\right)-f\left(a\right)}{b-a}$ to calculate the average rate of change of distance.

$\begin{array}{r}\frac{f\left(10\right)-f\left(0\right)}{10-0}=\frac{150-0}{10}=15\\ \end{array}$

Therefore, the average rate of change of distance over an interval $\left[0,10\right]$ is $15$.

Consider the second $10$ minutes interval.

The distance from a tree at $t=10$ is $150$ and at $t=20$ is $250$.

Use the formula role="math" localid="1648396762979" $\phantom{\rule{0ex}{0ex}}\frac{f\left(b\right)-f\left(a\right)}{b-a}$ to calculate the average rate of change of distance.

$\begin{array}{r}\frac{f\left(20\right)-f\left(10\right)}{20-10}=\frac{250-150}{10}\\ =\frac{100}{10}\\ =10\end{array}$

Therefore, the average rate of change of distance over an interval $\left[10,20\right]$ is $10$.

Consider the last $10$ minutes interval.

The distance from tree at $t=20$ is $250$ and at $t=30$ is $0$.

Use the formula role="math" localid="1648396865679" $\phantom{\rule{0ex}{0ex}}\frac{f\left(b\right)-f\left(a\right)}{b-a}$ to calculate the average rate of change of distance.

$\begin{array}{r}\frac{f\left(30\right)-f\left(20\right)}{30-20}=\frac{0-250}{10}\\ =\frac{-250}{10}\\ =-25\end{array}$

Therefore, the average rate of change of distance over an interval $\left[20,30\right]$ is $-25$.

The average rate of change of distance is maximum on the first 10 minutes interval is $15$ feet per second.

## Step 4.  (c)  The main objective is to determine the average velocity of the person L on an interval [5,10]. Explain the meaning of sign in average velocity in real world.

An approximated value of the distance from the tree at $t=5$ is $280$ and at $t=10$ is $150$. Use the formula role="math" localid="1648398177267" $\phantom{\rule{0ex}{0ex}}\frac{f\left(b\right)-f\left(a\right)}{b-a}$ to calculate the average rate of change of distance.

$\begin{array}{r}\frac{f\left(10\right)-f\left(5\right)}{10-5}=\frac{150-280}{5}\\ =\frac{-130}{5}\\ =-26\end{array}$

The negative value of average velocity indicates that the person moves towards the tree.

## Step 5.  (d)  The main objective is to determine the time at which the velocity of person L is zero and explain the significance of that time.

The slope of the distance function gives the velocity function.

From the graph, it is observed that the slope of the distance function is zero where the curve has maximum and minimum values.

It is observed that the slope of the function is zero at $t=4,t=12$ and $t=25$.

Thus, an approximated value of time $t$ is $t=4$ minutes, $t=12$ minutes and $t=25$ minutes at which the velocity is zero. It implies that person L changes the direction of the position.

## Step 6.  (e)   The main objective is to determine the time at which the velocity of person L is negative and explain the significance of that time.

Person L has negative velocity when the distance decreases.

From the graph, it is observed that the curve falls at $t=4$ minutes and at $t=25$ minutes. Therefore, the distance of the person decreases at $t=4$ minutes and $t=25$minutes. The negative value of velocity indicates that person L moves towards the tree. ### Want to see more solutions like these? 