Answer
a) $v(t)= 4t^{3}~-~6t^{2}~+~2t~-1$
$a(t)= 12t^{2}~-~12t~+~2$
b) $a(1) = 2\text{ m/s}^{-2}$
c) See graphs
Work Step by Step
a) The velocity is the first derivation of the function of the position of time, and the acceleration is the second derivation of the function of the position of time:
$v(t) = s^{'}(t) = 4t^{3}~-~6t^{2}~+~2t~-1$
$a(t) = s^{''}(t) = 12t^{2}~-~12t~+~2$
(b) We substitute $t = 1$ to $a(t)$:
$a(t) = s^{''}(t) = 12t^{2}~-~12t~+~2$
$a(2) = 12(1)^{2} - 12(1) +2=2\text{ m/s}^{-2}$
c) Graph the three functions on the same graph.
