## Calculus with Applications (10th Edition)

$\frac{28}{5}$
We are given $s(t)=5t^2-2t-7$ For $t = 3$ the instantaneous velocity is $\lim\limits_{b \to 3}\frac{s(b)-s(3)}{b-3}$ $s(b)=5b^2-2b-7$ $s(3)=32$ The instantaneous rate of change is then $\lim\limits_{b \to 3}\frac{5b^2-2b-7-32}{b-3}=\lim\limits_{b \to 3}\frac{(b-3)(b+\frac{13}{5})}{b-3}=\lim\limits_{b \to 3}(b+\frac{13}{5})=\frac{28}{5}$