Answer
(a)
$v(t)=4t^{3}-6t^{2}+2t-1$
$a(t)=12t^{2}-12t+2$
(b) a(1) = 2 s
(c) See graph.
Work Step by Step
$s(t)=t^{4}-2t^{3}+t^{2}-t$
(a)
Velocity is the derivative of position, so to get the velocity function in terms of t, we just need to take the derivative of s(t):
$s'(t) = v(t)=(4)t^{3}-2(3)t^{2}+2t-1$
$v(t)=4t^{3}-6t^{2}+2t-1.$
Acceleration is the derivative of velocity:
$v'(t) =a(t)=4(3)t^{2}-6(2)t+2-0$
$a(t)=12t^{2}-12t+2.$
(b) To find the acceleration at t = 1, just plug it into a(t):
$a(1)=12(1)^{2}-12(1)+2 = 2 s$
