Sketch the graph of $y = v(t)$. Indicate clearly the local maximum and write down its coordinates.

Use the substitution $u = {t^2}$ to find $\int {\frac{t}{{12 + {t^4}}}{\text{d}}t}$.

Find the exact distance travelled by particle $A$ between $t = 0$ and $t = 6$ seconds.

Give your answer in the form $k\arctan (b),{\text{ }}k,{\text{ }}b \in \mathbb{R}$.

Find the acceleration of particle B when $s = 0.1{\text{ m}}$.

(a)
    A1

A1 for correct shape and correct domain

$(1.41,{\text{ }}0.0884){\text{ }}\left( {\sqrt 2 ,{\text{ }}\frac{{\sqrt 2 }}{{16}}} \right)$     A1

[2 marks]

EITHER

$u = {t^2}$

$\frac{{{\text{d}}u}}{{{\text{d}}t}} = 2t$     A1

OR

$t = {u^{\frac{1}{2}}}$

$\frac{{{\text{d}}t}}{{{\text{d}}u}} = \frac{1}{2}{u^{ - \frac{1}{2}}}$     A1

THEN

$\int {\frac{t}{{12 + {t^4}}}{\text{d}}t = \frac{1}{2}\int {\frac{{{\text{d}}u}}{{12 + {u^2}}}} }$     M1

$= \frac{1}{{2\sqrt {12} }}\arctan \left( {\frac{u}{{\sqrt {12} }}} \right)( + c)$     M1

$= \frac{1}{{4\sqrt 3 }}\arctan \left( {\frac{{{t^2}}}{{2\sqrt 3 }}} \right)( + c)$ or equivalent     A1

[4 marks]

$\int_0^6 {\frac{t}{{12 + {t^4}}}{\text{d}}t}$     (M1)

$= \left[ {\frac{1}{{4\sqrt 3 }}\arctan \left( {\frac{{{t^2}}}{{2\sqrt 3 }}} \right)} \right]_0^6$     M1

$= \frac{1}{{4\sqrt 3 }}\left( {\arctan \left( {\frac{{36}}{{2\sqrt 3 }}} \right)} \right){\text{ }}\left( { = \frac{1}{{4\sqrt 3 }}\left( {\arctan \left( {\frac{{18}}{{\sqrt 3 }}} \right)} \right)} \right){\text{ (m)}}$     A1

Note:     Accept $\frac{{\sqrt 3 }}{{12}}\arctan \left( {6\sqrt 3 } \right)$ or equivalent.

[3 marks]

$\frac{{{\text{d}}v}}{{{\text{d}}s}} = \frac{1}{{2\sqrt {s(1 - s)} }}$     (A1)

$a = v\frac{{{\text{d}}v}}{{{\text{d}}s}}$

$a = \arcsin \left( {\sqrt s } \right) \times \frac{1}{{2\sqrt {s(1 - s)} }}$     (M1)

$a = \arcsin \left( {\sqrt {0.1} } \right) \times \frac{1}{{2\sqrt {0.1 \times 0.9} }}$

$a = 0.536{\text{ (m}}{{\text{s}}^{ - 2}})$     A1

[3 marks]