Sketch the graph of ${v_A} = {t^3} - 5{t^2} + 6t$ for $t \geqslant 0$, with ${v_A}$ on the vertical axis and t on the horizontal. Show on your sketch the local maximum and minimum points, and the intercepts with the t-axis.

Write down the times for which the velocity of the particle is increasing.

Write down the times for which the magnitude of the velocity of the particle is increasing.

At t = 0 the particle is at point O on the line.

Find an expression for the particle’s displacement, ${x_A}{\text{m}}$, from O at time t.

A second particle, B, moving along the same line, has position ${x_B}{\text{ m}}$, velocity ${v_B}{\text{ m}}{{\text{s}}^{ - 1}}$ and acceleration, ${a_B}{\text{ m}}{{\text{s}}^{ - 2}}$, where ${a_B} = - 2{v_B}$ for $t \geqslant 0$. At $t = 0,{\text{ }}{x_B} = 20$ and ${v_B} = - 20$.

Find an expression for ${v_B}$ in terms of t.

Find the value of t when the two particles meet.

     A1A1A1

Note: Award A1 for general shape, A1 for correct maximum and minimum, A1 for intercepts.

Note: Follow through applies to (b) and (c).

[3 marks]

$0 \leqslant t < 0.785,{\text{ }}\left( {{\text{or }}0 \leqslant t < \frac{{5 - \sqrt 7 }}{3}} \right)$     A1

(allow $t < 0.785$)

and $t > 2.55{\text{ }}\left( {{\text{or }}t > \frac{{5 + \sqrt 7 }}{3}} \right)$     A1

[2 marks]

$0 \leqslant t < 0.785,{\text{ }}\left( {{\text{or }}0 \leqslant t < \frac{{5 - \sqrt 7 }}{3}} \right)$     A1

(allow $t < 0.785$)

$2 < t < 2.55,{\text{ }}\left( {{\text{or }}2 < t < \frac{{5 + \sqrt 7 }}{3}} \right)$     A1

$t > 3$     A1

[3 marks]

position of A: ${x_A} = \int {{t^3} - 5{t^2} + 6t{\text{ d}}t}$     (M1)

${x_A} = \frac{1}{4}{t^4} - \frac{5}{3}{t^3} + 3{t^2}\,\,\,\,\,( + c)$     A1

when $t = 0,{\text{ }}{x_A} = 0$, so $c = 0$     R1

[3 marks]

$\frac{{{\text{d}}{v_B}}}{{{\text{d}}t}} = - 2{v_B} \Rightarrow \int {\frac{1}{{{v_B}}}{\text{d}}{v_B} = \int { - 2{\text{d}}t} }$     (M1)

$\ln \left| {{v_B}} \right| = - 2t + c$     (A1)

${v_B} = A{e^{ - 2t}}$     (M1)

${v_B} = - 20$ when t = 0 so ${v_B} = - 20{e^{ - 2t}}$     A1

[4 marks]

${x_B} = 10{e^{ - 2t}}( + c)$     (M1)(A1)

${x_B} = 20{\text{ when }}t = 0{\text{ so }}{x_B} = 10{e^{ - 2t}} + 10$     (M1)A1

meet when $\frac{1}{4}{t^4} - \frac{5}{3}{t^3} + 3{t^2} = 10{e^{ - 2t}} + 10$     (M1)

$t = 4.41(290 \ldots )$     A1

[6 marks]

Part (a) was generally well done, although correct accuracy was often a problem.

Parts (b) and (c) were also generally quite well done.

Parts (b) and (c) were also generally quite well done.

A variety of approaches were seen in part (d) and many candidates were able to obtain at least 2 out of 3. A number missed to consider the $+c$ , thereby losing the last mark.

Surprisingly few candidates were able to solve part (e) correctly. Very few could recognise the easy variable separable differential equation. As a consequence part (f) was frequently left.

Surprisingly few candidates were able to solve part (e) correctly. Very few could recognise the easy variable separable differential equation. As a consequence part (f) was frequently left.