Chapter 1.5, Problem 68E

(a)

To determine

To find:The function that represent the situation of the given equation.

Answer to Problem 68E

The function that represent the situation of the given equation is $s=-16t^2+72t+6\cdot 5$ .

Explanation of Solution

Given information:

The given equationis $s=-16t^2+v_{0}t+s_0$ .

The height is $s_0=6\cdot 5\text{feet}$ .

The initial velocity is $v_0=72\text{feet/s}$ .

Calculation:

Using the position $s=-16t^2+v_{0}t+s_0$ .

The function is.

  $s=-16t^2+72t+6\cdot 5$

Therefore, the function that represent the situation of the given equation is $s=-16t^2+72t+6\cdot 5$ .

(b)

To determine

To graph: For the given function.

Explanation of Solution

Given information:

The given function is $s=-16t^2+72t+6\cdot 5$ .

Graph:

The graph for the given function is shown in figure (1).

  

Figure (1)

Interpretation: Graph for the function $s=-16t^2+72t+6\cdot 5$ is shown in figure (1).

(c)

To determine

To find: The average rate of change of the function from $t_1$ to $t_2$ .

Answer to Problem 68E

The average rate of change of the function from $t_1$ to $t_2$ is $8$ .

Explanation of Solution

Given information:

The given equation is $s=-16t^2+72t+6\cdot 5$ .

The values are $t_1=0$ and $t_2=4$ .

Calculation:

Average rate of change of the function from $t_1$ to $t_2$ .

  $\frac{f\left(t_2\right)-f\left(t_1\right)}{t_2-t_1}$

Calculate the values at $t_1=0$ and $t_2=4$ .

  $\begin{array}{c}f\left(0\right)=-16{\left(0\right)}^2+72\left(0\right)+6\cdot 5\\ =38\cdot 5\\ f\left(4\right)=-16{\left(4\right)}^2+72\left(4\right)+6\cdot 5\\ =6\cdot 5\end{array}$

Calculate average rate of change of the function from $t_1$ to $t_2$ .

  $\begin{array}{c}=\frac{f\left(3\right)-f\left(0\right)}{4-0}\\ =\frac{38\cdot 5-6\cdot 5}{4}\\ =8\end{array}$

Therefore, the average rate of change of the function from $t_1$ to $t_2$ is $8$ .

(d)

To determine

To find: The behavior of slope of the secant line through $t_1$ to $t_2$ .

Answer to Problem 68E

The rate of change of the function represents that in one second the object travels $8$ feet.

Explanation of Solution

Given information:

The given equation is $s=-16t^2+72t+6\cdot 5$ .

The values are $t_1=0$ and $t_2=4$ .

Calculation:

The rate of change of the function represents that in one second the object travels $8$ feet.

Therefore, the rate of change of the function represents that in one second the object travels $8$ feet.

(e)

To determine

To find: The slope of the secant line through $t_1$ to $t_2$ .

Answer to Problem 68E

The slope of the secant line through $t_1$ to $t_2$ is $f\left(t\right)=8t+6\cdot 5$ .

Explanation of Solution

Given information:

The given equation is $s=-16t^2+72t+6\cdot 5$ .

Calculation:

Using the above value.

  $\frac{f\left(t_2\right)-f\left(t_1\right)}{t_2-t_1}=16$

The equation of secant line is.

  $\begin{array}{c}\left(f\left(t\right)-f\left(0\right)\right)=\frac{f\left(t_2\right)-f\left(t_1\right)}{t_2-t_1}\left(t-0\right)\\ \left(f\left(t\right)-6\cdot 5\right)=\frac{38\cdot 5-6\cdot 5}{4-0}\left(t-0\right)\\ f\left(t\right)=8t+6\cdot 5\end{array}$

Therefore, the slope of the secant line through $t_1$ to $t_2$ is $f\left(t\right)=8t+6\cdot 5$ .

(f)

To determine

To graph: For the secant line.

Explanation of Solution

Given information:

The secant line is $f\left(t\right)=8t+6\cdot 5$ .

Graph:

The graph for the secant line is shown in figure (1).

  

Figure (1)

Interpretation: Graph for the secant line $f\left(t\right)=8t+6\cdot 5$ is shown in figure (1).