Answer
The answers for the table in the given order:
$104 {ft\over {sec}}$
$97.6{ft\over {sec}}$
$96.16{ft\over {sec}}$
$96.016{ft\over {sec}}$
$96.0016{ft\over {sec}}$
The value the average speed approaches is $ 96 {ft\over {sec}}$.
Yes, it is reasonable.
Work Step by Step
The answers for the table in the given order:
$a=3$, $b=3.5$, $v_{avg}={{d(3.5)-d(3)}\over {3.5-3}}={{16(3.5)^2-16(3)^2}\over {3.5-3}}=104 {ft\over {sec}}$
$a=3$, $b=3.1$, $v_{avg}={{d(3.1)-d(3)}\over {3.1-3}}={{16(3.1)^2-16(3)^2}\over {3.1-3}}=97.6{ft\over {sec}}$
$a=3$, $b=3.01$, $v_{avg}={{d(3.01)-d(3)}\over {3.01-3}}={{16(3.01)^2-16(3)^2}\over {3.01-3}}=96.16{ft\over {sec}}$
$a=3$, $b=3.001$, $v_{avg}={{d(3.001)-d(3)}\over {3.001-3}}={{16(3.001)^2-16(3)^2}\over {3.001-3}}=96.016{ft\over {sec}}$
$a=3$, $b=3.0001$, $v_{avg}={{d(3.0001)-d(3)}\over {3.0001-3}}={{16(3.0001)^2-16(3)^2}\over {3.0001-3}}=96.0016{ft\over {sec}}$
it is reasonable because, as the change in distance ($d(b)-d(a)$) and change in time $(b-a)$ become infinitesimally small, meaning that the value $b$ approaches the value $a$, the result becomes more accurate. The table indeed shows this behavior.
