Answer
(i) $0 ft/s$
(ii) $ 1 ft/s$
(iii) $ 3ft/s$
(iv) $ 4ft/s$
b. $2 ft/s$
Work Step by Step
(i) $[4.8]$
$\frac{s(8)-s(4)}{8-4} = \frac{(7)-(7)}{4} = \frac{0}{4}= 0 ft/s$
(ii) $[6,8]$
$\frac{s(8)-s(6)}{8-6} = \frac{(7)-(5)}{2} = \frac{2}{2} = 1 ft/s$
(iii) $[8,10]$
$\frac{s(10)-s(8)}{10-8} = \frac{(13)-(7)}{2} = \frac{6}{2} = 3 ft/s$
(iv) $[8,12]$
$\frac{s(12)-s(8)}{12-8} = \frac{(23)-(7)}{4} = \frac{16}{4} = 4 ft/s$
b. Find the instantaneous velocity when $t = 8$.
$s' = \lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$
$s' = \lim\limits_{h \to 0} \frac{1(a+h)^{2} - 6(a+h) + 23 - (\frac{1}{2} a^{2} - 6a + 23)}{h}$
$s' = \lim\limits_{h \to 0} \frac{\frac{1}{2}a^{2} + ah + \frac{1}{2} h^{2} - 6a +6h + 23 - \frac{1}{2}a^{2} +6a - 23}{h}$
$s' = \lim\limits_{h \to 0} \frac{\frac{1}{2}h^{2} + ah + 6h}{h}$
Simplify:
$s' = \lim\limits_{h \to 0} \frac{h(\frac{1}{2}h + a - 6)}{h}$
$s' = \lim\limits_{h \to 0} \frac{1}{2}h + a - 6$
$s' = \frac{1}{2}(0) + a - 6$
$s' = a - 6$
$s'(8) = 8 - 6 = 2 ft/s$
