StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q. 51

Expert-verifiedFound in: Page 167

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 $[0,30]$ 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 $[20,30]$ 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.

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.

The average rate of change of a function $f\left(x\right)$ over a time interval $[a,b]$ 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 $[0,30]$.

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

Therefore, the average rate of change of the person L of distance over an interval $[0,30]$ 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 the same as that of at $t=0$.

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 $[0,10]$ 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 $[10,20]$ 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 $[20,30]$ is $-25$.

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

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.

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.

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.

94% of StudySmarter users get better grades.

Sign up for free