Answer
a)
i) $3$ $m/s$
ii) $2.75$ $m/s$
iii) $2.625$ $m/s$
iv) $2.525$ $m/s$
b) $2.5$ $m/s$
Work Step by Step
a)
$s$ = $s(t)$ = $1+2t+\frac{t^{2}}{4}$
the average velocity over the time interval [1,1+h] is
$v_{avg}$ = $\frac{s(1+h)-s(1)}{(1+h)-1}$ = $\frac{1+2(1+h)+\frac{(1+h)^{2}}{4}-\frac{13}{4}}{h}$ = $\frac{10+h}{4}$
so
i) [1,3], $h$ = $2$
$v_{avg}$ = $\frac{10+2}{4}$ = $3$ $m/s$
ii) [1,2], $h$ = $1$
$v_{avg}$ = $\frac{10+1}{4}$ = $2.75$ $m/s$
iii) [1,1.5], $h$ = $0.5$
$v_{avg}$ = $\frac{10+0.5}{4}$ = $2.625$ $m/s$
iv) [1,1.1], $h$ = $0.1$
$v_{avg}$ = $\frac{10+0.1}{4}$ = $2.525$ $m/s$
b)
when $t$ = $1$ the instantaneous velocity is
$\lim\limits_{h \to 0}{\frac{s(1+h)-s(1)}{h}}$ = $\lim\limits_{h \to 0}{\frac{10+h}{4}}$ = $\frac{10}{4}$ = $2.5$ $m/s$
