Chapter 1.5, Problem 70E

(a)

To determine

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

Answer to Problem 70E

The function that represent the situation of the given equation is $s=-16t^2+80$ .

Explanation of Solution

Given information:

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

The height is $s_0=80\text{feet}$ .

The initial velocity is $v_0=0\text{feet/s}$ .

Calculation:

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

The function is.

  $s=-16t^2+80$

Therefore, the function that represent the situation of the given equation is $s=-16t^2+80$ .

(b)

To determine

To graph: For the given function.

Explanation of Solution

Given information:

The given function is $s=-16t^2+80$ .

Graph:

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

  

Figure (1)

Interpretation: Graph for the function $s=-16t^2+80$ 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 70E

The average rate of change of the function from $t_1$ to $t_2$ is $-48$ .

Explanation of Solution

Given information:

The given equation is $s=-16t^2+64t+6$ .

The values are $t_1=1$ and $t_2=2$ .

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 $t_1=1$ and $t_2=2$ .

  $\begin{array}{c}f\left(2\right)=-16{\left(2\right)}^2+80\\ =16\\ f\left(1\right)=-16{\left(1\right)}^2+80\\ =64\end{array}$

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

  $\begin{array}{c}=\frac{f\left(3\right)-f\left(0\right)}{2-1}\\ =\frac{16-64}{1}\\ =-48\end{array}$

Therefore, the average rate of change of the function from $t_1$ to $t_2$ is $-48$ .

(d)

To determine

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

Answer to Problem 70E

The slope of the secant line through $t_1$ to $t_2$ is negative.

Explanation of Solution

Given information:

The given equation is $s=-16t^2+64t+6$ .

The values are $t_1=1$ and $t_2=2$ .

Calculation:

The average rate of change between any two points $\left(t_1,f\left(t_1\right)\right)$ and $\left(t_2,f\left(t_2\right)\right)$ .

Therefore, the slope of the secant line through $t_1$ to $t_2$ is negative.

(e)

To determine

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

Answer to Problem 70E

The slope of the secant line through $t_1$ to $t_2$ is $f\left(t\right)=48+16$ .

Explanation of Solution

Given information:

The given equation is $s=-16t^2+64t+6$ .

The values are $t_1=1$ and $t_2=2$ .

Calculation:

Using the above value.

  $\frac{f\left(t_2\right)-f\left(t_1\right)}{t_2-t_1}=-48$

The equation of secant line is.

  $\begin{array}{c}\left(f\left(t\right)-f\left(1\right)\right)=\frac{f\left(t_2\right)-f\left(t_1\right)}{t_2-t_1}\left(t-1\right)\\ \left(f\left(t\right)-64\right)=-48\left(t-1\right)\\ f\left(t\right)=48+16\end{array}$

Therefore, the slope of the secant line through $t_1$ to $t_2$ is $f\left(t\right)=48+16$ .

(f)

To determine

To graph: For the secant line.

Explanation of Solution

Given information:

The secant line is $f\left(t\right)=48+16$ .

Graph:

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

  

Figure (1)

Interpretation: Graph for the secant line $f\left(t\right)=48+16$ is shown in figure (1).