Q12 (b)

Criteria	Marks
Provides correct solution	3
Obtains the displacement of the particle at time $t$, or equivalent merit	2
Obtains the position of the particle at time $t$ OR finds the time at which the velocity of the particle is a maximum	1

Sample answer:
$\ddot x=12-6t$
$t=0,\quad v=0,\quad x=0$

$$\frac{dv}{dt}=12-6t,\qquad v=\int(12-6t)\,dt=12t-3t^2+c$$

$t=0,\ v=0\Rightarrow c=0,\qquad v=12t-3t^2$

$$\frac{dx}{dt}=12t-3t^2,\qquad x=6t^2-t^3+c$$

$t=0,\ v=0\Rightarrow c=0,\qquad x=6t^2-t^3$

Maximum velocity
$v=12t-3t^2$
$v$ is maximum when $t=2$.
$x=6\times2^2-2^3=16$
Position at maximum velocity is 16 units to right of origin.