Chapter 2.6, Problem 16E

(a)

To determine

To find: The average velocity of the particle at the given time intervals.

Answer to Problem 16E

The average velocity of the particle,

(i) At the interval [4, 8] is $\underset{\_}{0\text{ ft/s}}$.

(ii) At the interval [6, 8] is $\underset{\_}{1\text{ ft/s}}$.

(iii) At the interval [8, 10] is $\underset{\_}{3\text{ ft/s}}$.

(iv) At the interval [8, 12] is $\underset{\_}{4\text{ ft/s}}$.

Explanation of Solution

Given:

The displacement (in feet) of a particle moving in a straight line is $s=\frac{1}{2}{t}^2-6t+23$, where t is measured in seconds.

Formula used:

The average velocity over the time interval $\left[a,a+h\right]$ is,

$\frac{f\left(a+h\right)-f\left(a\right)}{h}$ (1)

Calculation:

Obtain the average velocity of the particle over the time interval $\left[a,a+h\right]$.

Take the position function $s\left(t\right)=f\left(t\right)$ and use equation (1) to compute the average velocity over the time interval $\left[a,a+h\right]$ as follows.

$\begin{array}{c}\text{Average velocity}=\frac{f\left(a+h\right)-f\left(a\right)}{h}\\ =\frac{\left(\frac{1}{2}{\left(a+h\right)}^2-6\left(a+h\right)+23\right)-\left(\frac{1}{2}{\left(a\right)}^2-6\left(a\right)+23\right)}{h}\\ =\frac{\left(\frac{1}{2}\left(a^2+h^2+2ha\right)-6a-6h+23\right)-\left(\frac{1}{2}{a}^2-6a+23\right)}{h}\\ =\frac{\frac{1}{2}{a}^2+\frac{1}{2}{h}^2+ha-6a-6h+23-\frac{1}{2}{a}^2+6a-23}{h}\end{array}$

Simplify further and obtain the value of average velocity as follows.

$\begin{array}{c}\text{Average velocity}=\frac{\frac{1}{2}{h}^2+ha-6h}{h}\\ =\frac{h\left(\frac{1}{2}h+a-6\right)}{h}\\ =\left(a+\frac{1}{2}h-6\right)\end{array}$

Thus, the average velocity of the particle over the time interval $\left[a,a+h\right]$ is $\underset{\_}{\left(a+\frac{1}{2}h-6\right)\text{ m/s}}$.

Section (i)

Obtain the average velocity of the particle over the time interval $\left[4,8\right]$.

Consider $a=4$ and $a+h=8$.

$\begin{array}{c}h=8-a\\ =8-\left(4\right)\\ =4\end{array}$

Substitute $a=4$ and $h=4$ in the average velocity,

$\begin{array}{c}\text{Average velocity}=a+\frac{1}{2}h-6\\ =\left(4\right)+\frac{1}{2}\left(4\right)-6\\ =-2+2\\ =0\end{array}$

Thus, the average velocity of the particle over the time interval $\left[4,8\right]$ is $0\text{ ft/s}$.

Section (ii)

Obtain the average velocity of the particle over the time interval $\left[6,8\right]$.

Consider $a=6$ and $a+h=8$.

$\begin{array}{c}h=8-a\\ =8-\left(6\right)\\ =2\end{array}$

Substitute $a=6$ and $h=2$ in the average velocity,

$\begin{array}{c}\text{Average velocity}=a+\frac{1}{2}h-6\\ =\left(6\right)+\frac{1}{2}\left(2\right)-6\\ =1\end{array}$

Thus, the average velocity of the particle over the time interval $\left[6,8\right]$ is $1\text{ ft/s}$.

Section (iii)

Obtain the average velocity of the particle over the time interval $\left[8,10\right]$.

Consider $a=8$ and $a+h=10$.

$\begin{array}{c}h=10-a\\ =10-\left(8\right)\\ =2\end{array}$

Substitute $a=8$ and $h=2$ in the average velocity,

$\begin{array}{c}\text{Average velocity}=a+\frac{1}{2}h-6\\ =\left(8\right)+\frac{1}{2}\left(2\right)-6\\ =2+1\\ =3\end{array}$

Thus, the average velocity of the particle over the time interval $\left[8,10\right]$ is $3\text{ ft/s}$  .

Section (iv)

Obtain the average velocity of the particle over the time interval $\left[8,12\right]$.

Take $a=8$ and $a+h=12$, so that

$\begin{array}{c}h=12-a\\ =12-\left(8\right)\\ =4\end{array}$

Substitute $a=8$ and $h=4$ in the average velocity,

$\begin{array}{c}\text{Average velocity}=a+\frac{1}{2}h-6\\ =\left(8\right)+\frac{1}{2}\left(4\right)-6\\ =2+2\\ =4\end{array}$

Thus, the average velocity of the particle over the time interval $\left[8,12\right]$ is $4\text{ ft/s}$.

(b)

To determine

To find: The instantaneous velocity at $t=8$ seconds.

Answer to Problem 16E

The instantaneous velocity when $t=8$ is $\underset{\_}{2\text{ ft/s}}$.

Explanation of Solution

Formula used:

The derivative $v\left(a\right)$ is the instantaneous velocity of the particle at time $t=a$.

$v\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ (2)

Calculation:

From part (a), $\frac{f\left(a+h\right)-f\left(a\right)}{h}=a+\frac{1}{2}h-6$.

Use equation (2) to obtain the instantaneous velocity,

$\begin{array}{c}v\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}\\ =\underset{h\to 0}{\lim }\left(a+\frac{1}{2}h-6\right)\\ =a+\frac{1}{2}\left(0\right)-6\\ =a-6\end{array}$

Therefore, the instantaneous velocity at $t=a$ is $v\left(a\right)=a-6$.

Substitute $a=8$ in $v\left(a\right)=a-6$,

$\begin{array}{c}v\left(8\right)=8-6\\ =2\end{array}$

Thus, the instantaneous velocity at $t=8$ is $2\text{ ft/s}$.

(c)

To determine

To sketch: The graph of the function $s\left(t\right)$, secant lines and the tangent line.

Explanation of Solution

Given:

The equation of the position function is $s=\frac{1}{2}{t}^2-6t+23$.

The secant lines whose slopes are the average velocities in part (a).

The tangent line whose slope is the instantaneous velocity in part (b).

Graph:

Use the online graphing calculator to draw the graph of the function $s\left(t\right)$, secant lines and tangent line as shown below in Figure 1.

From the Figure 1, it is observed that the tangent line touches the curve at $t=8$. The secant lines passes through the curve at the respected given time intervals.