Chapter 1.5, Problem 69E

(a)

To determine

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

Answer to Problem 69E

The function that represent the situation of the given equation is $s=-16t^2+120t$ .

Explanation of Solution

Given information:

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

The height is $s_0=0\text{feet}$ .

The initial velocity is $v_0=120\text{feet/s}$ .

Calculation:

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

The function is.

  $s=-16t^2+120t$

Therefore, the function that represent the situation of the given equation is $s=-16t^2+120t$ .

(b)

To determine

To graph: For the given function.

Explanation of Solution

Given information:

The given function is $s=-16t^2+120t$ .

Graph:

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

  

Figure (1)

Interpretation: Graph for the function $s=-16t^2+120t$ 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 69E

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 $s=-16t^2+64t+6$ .

The values are $t_1=3$ and $t_2=5$ .

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 $t_1=3$ and $t_2=5$ .

  $\begin{array}{c}f\left(3\right)=-16{\left(3\right)}^2+120\left(3\right)\\ =216\\ f\left(5\right)=-16{\left(5\right)}^2+120\left(5\right)\\ =200\end{array}$

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

  $\begin{array}{c}=\frac{f\left(5\right)-f\left(3\right)}{5-3}\\ =\frac{200-216}{2}\\ =-8\end{array}$

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 69E

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 $t_1=3$ and $t_2=5$ .

Calculation:

The average rate of change between any two points $\left(t_1,f\left(t_1\right)\right)$ and $\left(t_2,f\left(t_2\right)\right)$ is the slope of the line secant line is between two points $\left(3,216\right)$ and $\left(5,200\right)$ .

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 69E

The slope of the secant line through $t_1$ to $t_2$ is $f\left(t\right)=-8t+240$ .

Explanation of Solution

Given information:

The given equation is $f\left(t\right)=16t+6$ .

The values are $t_1=3$ and $t_2=5$ .

Calculation:

Using the above value.

  $\frac{f\left(t_2\right)-f\left(t_1\right)}{t_2-t_1}=-8$

The equation of secant line is.

  $\begin{array}{c}\left(f\left(t\right)-f\left(0\right)\right)=\frac{f\left(t_2\right)-f\left(t_1\right)}{t_2-t_1}\left(t-3\right)\\ \left(f\left(t\right)-216\right)=-8\left(t-3\right)\\ f\left(t\right)=-8t+240\end{array}$

Therefore, the slope of the secant line through $t_1$ to $t_2$ is $f\left(t\right)=-8t+240$ .

(f)

To determine

To graph: For the secant line.

Explanation of Solution

Given information:

The secant line is $f\left(t\right)=-8t+240$ .

Graph:

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

  

Figure (1)

Interpretation: Graph for the secant line $f\left(t\right)=-8t+240$ is shown in figure (1).