problem. 4 : the position function of a ball thrown into the air with a velocity of 23 ft/sec and initial height of 1 foot is given by the function $s(t) = -16t^2 + 23t + 1$ where $s(t)$ is the height of the ball above the ground after $t$ seconds. find the average velocity of the ball on the interval starting with $t = 4$ to the time 0.5 seconds later. $v_{ave} = -113$
problem. 4.1 : now write an expression for the average velocity of the ball on the time interval from $t = 4$ to $h$ seconds later. that is, find the expression for the average velocity on the interval $4, 4 + h$ where $h \
eq 0$.
problem. 5 : find the average velocity of $s(t) = -10t^2 + 8t + 4$ from $t = 4$ to $t = 4 + h$.
$v_{ave} = $
problem. 6 : use algebra to find the following limit
$\lim_{h \to 0} \frac{15\left((h - 8)^2 - 64\
ight)}{h} = $

Question

Accepted Answer

### Problem 4.1
# Explanation:
## Step1: Recall average velocity formula
The formula for average velocity $ V_{ave} $ over the interval $[a, b]$ is $ V_{ave}=\frac{s(b)-s(a)}{b - a} $. Here, $ a = 4 $ and $ b=4 + h $, so we need to find $ s(4 + h) $ and $ s(4) $, then substitute into the formula.

## Step2: Find $ s(4 + h) $
Given $ s(t)=- 16t^{2}+23t + 1 $, substitute $ t = 4+h $:
$$
\begin{align*}
s(4 + h)&=-16(4 + h)^{2}+23(4 + h)+1\
&=-16(16 + 8h+h^{2})+92 + 23h+1\
&=-256-128h-16h^{2}+92 + 23h + 1\
&=-16h^{2}-105h-163
\end{align*}
$$

## Step3: Find $ s(4) $
Substitute $ t = 4 $ into $ s(t) $:
$$
\begin{align*}
s(4)&=-16(4)^{2}+23(4)+1\
&=-256 + 92+1\
&=-163
\end{align*}
$$

## Step4: Substitute into average velocity formula
$$
\begin{align*}
V_{ave}&=\frac{s(4 + h)-s(4)}{(4 + h)-4}\
&=\frac{(-16h^{2}-105h - 163)-(-163)}{h}\
&=\frac{-16h^{2}-105h}{h}\
&=-16h-105\quad(h
eq0)
\end{align*}
$$
# Answer:
$ -16h - 105 $

### Problem 5
# Explanation:
## Step1: Recall average velocity formula
The formula for average velocity $ V_{ave} $ over $[a,b]$ is $ V_{ave}=\frac{s(b)-s(a)}{b - a} $. Here, $ a = 4 $, $ b = 4+h $, so find $ s(4 + h) $ and $ s(4) $.

## Step2: Find $ s(4 + h) $
Given $ s(t)=-10t^{2}+8t + 4 $, substitute $ t = 4 + h $:
$$
\begin{align*}
s(4 + h)&=-10(4 + h)^{2}+8(4 + h)+4\
&=-10(16 + 8h+h^{2})+32 + 8h+4\
&=-160-80h-10h^{2}+32 + 8h + 4\
&=-10h^{2}-72h-124
\end{align*}
$$

## Step3: Find $ s(4) $
Substitute $ t = 4 $ into $ s(t) $:
$$
\begin{align*}
s(4)&=-10(4)^{2}+8(4)+4\
&=-160 + 32+4\
&=-124
\end{align*}
$$

## Step4: Substitute into average velocity formula
$$
\begin{align*}
V_{ave}&=\frac{s(4 + h)-s(4)}{(4 + h)-4}\
&=\frac{(-10h^{2}-72h-124)-(-124)}{h}\
&=\frac{-10h^{2}-72h}{h}\
&=-10h-72\quad(h
eq0)
\end{align*}
$$
# Answer:
$ -10h - 72 $

### Problem 6
# Explanation:
## Step1: Factor the numerator
Notice that $ (h - 8)^{2}-64=(h - 8)^{2}-8^{2} $, which is a difference of squares. Using $ a^{2}-b^{2}=(a - b)(a + b) $, where $ a=h - 8 $ and $ b = 8 $:
$$
\begin{align*}
(h - 8)^{2}-64&=(h - 8 - 8)(h - 8+8)\
&=(h - 16)h
\end{align*}
$$

## Step2: Simplify the limit expression
The limit is $ \lim_{h
ightarrow0}\frac{15((h - 8)^{2}-64)}{h} $. Substitute the factored form of the numerator:
$$
\begin{align*}
\lim_{h
ightarrow0}\frac{15\times h(h - 16)}{h}&=\lim_{h
ightarrow0}15(h - 16)
\end{align*}
$$
(We can cancel $ h $ since $ h
ightarrow0 $ but $ h
eq0 $ when taking the limit)

## Step3: Evaluate the limit
Substitute $ h = 0 $ into $ 15(h - 16) $:
$$
15(0 - 16)=-240
$$
# Answer:
$ -240 $

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QUESTION IMAGE

Question

Explanation:

Problem 4.1

Step1: Recall average velocity formula

Step2: Find \( s(4 + h) \)

Step3: Find \( s(4) \)

Step4: Substitute into average velocity formula

Step1: Recall average velocity formula

Step2: Find \( s(4 + h) \)

Step3: Find \( s(4) \)

Step4: Substitute into average velocity formula

Step1: Factor the numerator

Step2: Simplify the limit expression

Step3: Evaluate the limit

Answer:

Problem 5