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.