calculus i - exam 1 review - fall 2025\n1. an object moves in a straight line. the position in feet after…

Explanation:

Step1: Recall average - velocity formula

The average - velocity formula between $t = a$ and $t = b$ for a position function $s(t)$ is $v_{avg}=\frac{s(b)-s(a)}{b - a}$. Here, $a = 1$, $b = 3$, and $s(t)=5-\frac{8}{t + 1}$. First, find $s(1)$ and $s(3)$: [ \begin{align*} s(1)&=5-\frac{8}{1 + 1}=5 - 4=1\ s(3)&=5-\frac{8}{3 + 1}=5 - 2=3 \end{align*} ] Then, calculate the average - velocity: [ v_{avg}=\frac{s(3)-s(1)}{3 - 1}=\frac{3 - 1}{2}=1 ]

Step2: Recall instantaneous - velocity formula

The instantaneous - velocity function $v(t)$ is the derivative of the position function $s(t)$. We can also find the instantaneous velocity at $t = c$ using the limit formula $v(c)=\lim_{h\rightarrow0}\frac{s(c + h)-s(c)}{h}$. Here, $c = 3$, $s(t)=5-\frac{8}{t + 1}$, so $s(3 + h)=5-\frac{8}{(3 + h)+1}=5-\frac{8}{h + 4}$ and $s(3)=3$. [ \begin{align*} v(3)&=\lim_{h\rightarrow0}\frac{s(3 + h)-s(3)}{h}\ &=\lim_{h\rightarrow0}\frac{\left(5-\frac{8}{h + 4}\right)-3}{h}\ &=\lim_{h\rightarrow0}\frac{2-\frac{8}{h + 4}}{h}\ &=\lim_{h\rightarrow0}\frac{\frac{2(h + 4)-8}{h + 4}}{h}\ &=\lim_{h\rightarrow0}\frac{2h+8 - 8}{h(h + 4)}\ &=\lim_{h\rightarrow0}\frac{2h}{h(h + 4)}\ &=\lim_{h\rightarrow0}\frac{2}{h + 4}=\frac{2}{4}=\frac{1}{2} \end{align*} ]

Answer:

a) The average velocity between $t = 1$ and $t = 3$ is $1$ foot per second. b) The instantaneous velocity at $t = 3$ is $\frac{1}{2}$ foot per second.