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]