Answer
$\frac{28}{5}$
Work Step by Step
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
$\lim\limits_{b \to 3}\frac{5b^2-2b-7-32}{b-3}=\lim\limits_{b \to 3}\frac{(b-3)(b+\frac{13}{5})}{b-3}=\lim\limits_{b \to 3}(b+\frac{13}{5})=\frac{28}{5}$
