week 2 group project\nthe police department has a machine that makes a graph for each car that drives on a…

Explanation:

Step1: Recall average - velocity formula

The average velocity $v_{avg}$ of an object over the interval $[a,b]$ for a position - function $s(t)$ is given by $v_{avg}=\frac{s(b)-s(a)}{b - a}$.

Step2: Calculate average velocity for interval [2,3] for cars (A), (B), and (C)

Let $s_A(t)$, $s_B(t)$, and $s_C(t)$ be the position - functions of cars (A), (B), and (C) respectively. Then $v_{A,[2,3]}=\frac{s_A(3)-s_A(2)}{3 - 2}=s_A(3)-s_A(2)$, $v_{B,[2,3]}=s_B(3)-s_B(2)$, $v_{C,[2,3]}=s_C(3)-s_C(2)$.

Step3: Calculate average velocity for interval [2,2.5] for cars (A), (B), and (C)

$v_{A,[2,2.5]}=\frac{s_A(2.5)-s_A(2)}{2.5 - 2}=\frac{s_A(2.5)-s_A(2)}{0.5}=2(s_A(2.5)-s_A(2))$, $v_{B,[2,2.5]}=2(s_B(2.5)-s_B(2))$, $v_{C,[2,2.5]}=2(s_C(2.5)-s_C(2))$. The calculation over the interval $[2,2.5]$ is closer to the instantaneous velocity at $t = 2$ minutes because the interval is smaller.

For question 2: Choose a smaller interval, say $[2,2.1]$. Calculate $v_{A,[2,2.1]}=\frac{s_A(2.1)-s_A(2)}{2.1 - 2}=10(s_A(2.1)-s_A(2))$, $v_{B,[2,2.1]}=10(s_B(2.1)-s_B(2))$, $v_{C,[2,2.1]}=10(s_C(2.1)-s_C(2))$. As $h$ gets smaller, the average velocity over $[2,2 + h]$ approaches the instantaneous velocity at $t = 2$.

For question 3: A position - function $s(t)$ with a relatively linear behavior over the interval $[2,3]$ and more curvature over $[2,2.5]$ would satisfy the condition. For example, a linear function $s(t)=mt + c$ where the slope $m$ is constant over $[2,3]$ and changes more rapidly over $[2,2.5]$.

For question 4:

Part a

1. Simplify the limit $\lim_{x\rightarrow3}\frac{x^{2}-9}{x - 3}$.
   • Factor the numerator: $x^{2}-9=(x + 3)(x - 3)$.
   • Then $\lim_{x\rightarrow3}\frac{x^{2}-9}{x - 3}=\lim_{x\rightarrow3}\frac{(x + 3)(x - 3)}{x - 3}=\lim_{x\rightarrow3}(x + 3)=6$. So the statement "The value of $\lim_{x\rightarrow3}\frac{x^{2}-9}{x - 3}$ does not exist" is false.

Part b

1. Consider a function like $f(x)=\frac{x^{2}-9}{x - 3}$ for $x\neq3$ and undefined at $x = 3$.
   • $\lim_{x\rightarrow3}f(x)=6$, but $f(3)$ is undefined. So the statement "The value of $\lim_{x\rightarrow a}f(x)$ can always be found by computing $f(a)$" is false.

Part c

1. Consider the function $f(x)=\frac{x^{2}-9}{x - 3}$ again.
   • $f(3)$ is undefined, but $\lim_{x\rightarrow3}f(x)=6$. So the statement "The value of $\lim_{x\rightarrow a}f(x)$ does not exist if $f(a)$ is undefined" is false.

Answer:

1. For cars (A), (B), and (C), calculate average velocities using $v_{avg}=\frac{s(b)-s(a)}{b - a}$ for intervals $[2,3]$ and $[2,2.5]$. The $[2,2.5]$ calculation is closer to the instantaneous velocity at $t = 2$.
2. For cars (A), (B), and (C), calculate average velocity over $[2,2.1]$. As $h$ in $[2,2 + h]$ gets smaller, it approaches the instantaneous velocity at $t = 2$.
3. Create a position - function with relatively linear behavior over $[2,3]$ and more curvature over $[2,2.5]$.
4. • a. False. $\lim_{x\rightarrow3}\frac{x^{2}-9}{x - 3}=6$.
   • b. False. Example: $f(x)=\frac{x^{2}-9}{x - 3}$, $\lim_{x\rightarrow3}f(x)=6$ but $f(3)$ is undefined.
   • c. False. Example: $f(x)=\frac{x^{2}-9}{x - 3}$, $f(3)$ is undefined but $\lim_{x\rightarrow3}f(x)=6$.