An object travels on a line and has displacement function over time $s(t)=t^3+t^2+t^2+5,$

The object's avergge velocity between $t=1$ and $t=3$ can be calculated as

$=\frac{2(3)-(l)}{3}$

$=\frac{(3+2+2+3)-(2^2+2^2+2+5)}{2}$

$=\frac{N-8}{2}=\frac{38}{2}=18.$

Because $v(t)=s^{\text{'}}(t)=3t^2+2t+1,$ we may compute this way instead $--$

$\frac{?}{\text{b}\text{a}\text{r}}(D)=\frac{f{x}^2+2x+1c}{3-1}=\frac{36}{8}=18.$

Tue

False