Find the velocity of the particle when $t = 1$.

Find the value of $t$ for which the particle is at rest.

Find the total distance the particle travels during the first three seconds.

Show that the acceleration of the particle is given by $a = 6t{({t^2} - 4)^2}$.

Find all possible values of $t$ for which the velocity and acceleration are both positive or both negative.

substituting $t = 1$ into $v$     (M1)

eg     $v(1),{\text{ }}{\left( {{1^2} - 4} \right)^3}$

velocity $=  - 27{\text{ }}\left( {{\text{m}}{{\text{s}}^{ - 1}}} \right)$     A1     N2

[2 marks]

valid reasoning     (R1)

eg     $v = 0,{\text{ }}{\left( {{t^2} - 4} \right)^3} = 0$

correct working     (A1)

eg     ${t^2} - 4 = 0,{\text{ }}t =  \pm 2$, sketch

$t = 2$     A1     N2

[3 marks]

correct integral expression for distance     (A1)

eg     $\int_0^3 {\left| v \right|,{\text{ }}\int {\left| {{{\left( {{t^2} - 4} \right)}^3}} \right|,{\text{ }} - \int_0^2 {v{\text{d}}t + \int_2^3 {v{\text{d}}t} } } }$,

$\int_0^2 {{{\left( {4 - {t^2}} \right)}^3}{\text{d}}t + \int_2^3 {{{\left( {{t^2} - 4} \right)}^3}{\text{d}}t} }$ (do not accept $\int_0^3 {v{\text{d}}t}$)

$86.2571$

${\text{distance}} = 86.3{\text{ (m)}}$     A2     N3

[3 marks]

evidence of differentiating velocity     (M1)

eg     $v'(t)$

$a = 3{\left( {{t^2} - 4} \right)^2}(2t)$     A2

$a = 6t{\left( {{t^2} - 4} \right)^2}$     AG     N0

[3 marks]

METHOD 1

valid approach     M1

eg     graphs of $v$ and $a$

correct working     (A1)

eg     areas of same sign indicated on graph

$2 < t \leqslant 3$   (accept $t > 2$)     A2     N2

METHOD 2

recognizing that $a \geqslant 0$ (accept $a$ is always positive) (seen anywhere)     R1

recognizing that $v$ is positive when $t > 2$ (seen anywhere)     (R1)

$2 < t \leqslant 3$   (accept $t > 2$)     A2     N2

[4 marks]