| Date | May 2016 | Marks available | 1 | Reference code | 16M.2.hl.TZ1.3 |
| Level | HL only | Paper | 2 | Time zone | TZ1 |
| Command term | Find | Question number | 3 | Adapted from | N/A |
Question
The displacement, \(s\), in metres, of a particle \(t\) seconds after it passes through the origin is given by the expression \(s = \ln (2 - {e^{ - t}}),{\text{ }}t \geqslant 0\).
Find an expression for the velocity, \(v\), of the particle at time \(t\).
Find an expression for the acceleration, \(a\), of the particle at time \(t\).
Find the acceleration of the particle at time \(t = 0\).
Markscheme
\(v - \frac{{{\text{d}}s}}{{{\text{d}}t}} = \frac{{{{\text{e}}^{ - t}}}}{{2 - {{\text{e}}^{ - t}}}}{\text{ }}\left( { = \frac{1}{{2{{\text{e}}^t} - 1}}{\text{ or }} - 1 + \frac{2}{{2 - {{\text{e}}^{ - t}}}}} \right)\) M1A1
[2 marks]
\(a = \frac{{{{\text{d}}^2}s}}{{{\text{d}}{t^2}}} = \frac{{ - {{\text{e}}^{ - t}}(2 - {{\text{e}}^{ - t}}{\text{)}} - {{\text{e}}^{ - t}} \times {{\text{e}}^{ - t}}}}{{{{(2 - {{\text{e}}^{ - t}})}^2}}}{\text{ }}\left( { = \frac{{ - 2{{\text{e}}^{ - t}}}}{{{{(2 - {{\text{e}}^{ - t}})}^2}}}} \right)\) M1A1
Note: If simplified in part (a) award (M1)A1 for \(a = \frac{{{{\text{d}}^2}s}}{{{\text{d}}{t^2}}} = \frac{{ - 2{{\text{e}}^t}}}{{{{(2{{\text{e}}^t} - 1)}^2}}}\).
Note: Award M1A1 for \(a = - {{\text{e}}^{ - t}}{(2 - {{\text{e}}^{ - t}})^{ - 2}}({{\text{e}}^{ - t}}) - {{\text{e}}^{ - t}}{(2 - {{\text{e}}^{ - t}})^{ - 1}}\).
[2 marks]
\(a = - 2{\text{ }}({\text{m}}{{\text{s}}^{ - 2}})\) A1
[1 mark]
Examiners report
Mostly well done. There were a few sign errors but most candidates were correctly applying the quotient or chain rules.
Mostly well done. There were a few sign errors but most candidates were correctly applying the quotient or chain rules.
Mostly well done. There were a few sign errors but most candidates were correctly applying the quotient or chain rules.
