| Date | May 2012 | Marks available | 3 | Reference code | 12M.2.sl.TZ2.5 |
| Level | SL only | Paper | 2 | Time zone | TZ2 |
| Command term | Find | Question number | 5 | Adapted from | N/A |
Question
A particle moves in a straight line with velocity \(v = 12t - 2{t^3} - 1\) , for \(t \ge 0\) , where v is in centimetres per second and t is in seconds.
Find the acceleration of the particle after 2.7 seconds.
Find the displacement of the particle after 1.3 seconds.
Markscheme
recognizing that acceleration is the derivative of velocity (seen anywhere) (R1)
e.g. \(a = \frac{{{{\rm{d}}^2}s}}{{{\rm{d}}{t^2}}},v',12 - 6{t^2}\)
correctly substituting 2.7 into their expression for a (not into v) (A1)
e.g. \(s''(2.7)\)
\({\text{acceleration}} = - 31.74\) (exact), \( - 31.7\) A1 N3
[3 marks]
recognizing that displacement is the integral of velocity R1
e.g. \(s = \int v \)
correctly substituting 1.3 (A1)
e.g. \(\int_0^{1.3} {v{\rm{d}}t} \)
\({\text{displacement}} = 7.41195\) (exact), \(7.41\) (cm) A1 N2
[3 marks]
Examiners report
This question was well answered by many candidates, although there were some who did not recognize the relationship between velocity, acceleration and displacement. Many of them substituted into the original expression given for the velocity, losing most of the marks. Very few appear to have used their GDC for the integration.
This question was well answered by many candidates, although there were some who did not recognize the relationship between velocity, acceleration and displacement. Many of them substituted into the original expression given for the velocity, losing most of the marks. Very few appear to have used their GDC for the integration.
