Date | May 2013 | Marks available | 2 | Reference code | 13M.2.hl.TZ1.12 |
Level | HL only | Paper | 2 | Time zone | TZ1 |
Command term | Write down | Question number | 12 | Adapted from | N/A |
Question
A particle, A, is moving along a straight line. The velocity, \({v_A}{\text{ m}}{{\text{s}}^{ - 1}}\), of A t seconds after its motion begins is given by
\[{v_A} = {t^3} - 5{t^2} + 6t.\]
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.
Markscheme
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]
Examiners report
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.