Kinematics

$v = \dfrac{ds}{dt} = 8t - 3$, and at $t = 2$ this is $16 - 3 = 13$.

$a = \dfrac{dv}{dt} = 6$. It is constant — the $t$ disappears when $6t$ is differentiated.

$v = 3t^2 - 6t = 3t(t - 2)$, which is zero at $t = 0$ and $t = 2$.

$s = \displaystyle\int (2t+1)\,dt = t^2 + t + c$, and $s = 0$ at $t = 0$ gives $c = 0$. Then $s(3) = 9 + 3 = 12$.

The area is a triangle plus a rectangle: $\tfrac12(5)(20) = 50$ and $(9-5)(20) = 80$, giving $130$ m.

$s = t^3 - 6t^2 + 9t$. The velocity is zero at $t = 1$ and $t = 3$. Since $s(0)=0$, $s(1)=4$, $s(3)=0$ and $s(4)=4$, the three legs measure 4 m each, giving 12 m in total.