Question

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} = $

Explanation:

Problem 4.1

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

\[

$$\begin{align*} V_{ave}&=\frac{s(4 + h)-s(4)}{(4 + h)-4}\\ &=\frac{(-16h^{2}-105h - 163)-(-163)}{h}\\ &=\frac{-16h^{2}-105h}{h}\\ &=-16h-105\quad(h eq0) \end{align*}$$

\]

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

\[

$$\begin{align*} V_{ave}&=\frac{s(4 + h)-s(4)}{(4 + h)-4}\\ &=\frac{(-10h^{2}-72h-124)-(-124)}{h}\\ &=\frac{-10h^{2}-72h}{h}\\ &=-10h-72\quad(h eq0) \end{align*}$$

\]

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:

\( -16h - 105 \)

Problem 5