Date | November 2008 | Marks available | 5 | Reference code | 08N.1.sl.TZ0.9 |
Level | SL only | Paper | 1 | Time zone | TZ0 |
Command term | Find | Question number | 9 | Adapted from | N/A |
Question
The acceleration, \(a{\text{ m}}{{\text{s}}^{ - 2}}\), of a particle at time t seconds is given by \(a = 2t + \cos t\) .
Find the acceleration of the particle at \(t = 0\) .
Find the velocity, v, at time t, given that the initial velocity of the particle is \({\text{m}}{{\text{s}}^{ - 1}}\) .
Find \(\int_0^3 {v{\rm{d}}t} \) , giving your answer in the form \(p - q\cos 3\) .
What information does the answer to part (c) give about the motion of the particle?
Markscheme
substituting \(t = 0\) (M1)
e.g. \(a(0) = 0 + \cos 0\)
\(a(0) = 1\) A1 N2
[2 marks]
evidence of integrating the acceleration function (M1)
e.g. \(\int {(2t + \cos t){\text{d}}t} \)
correct expression \({t^2} + \sin t + c\) A1A1
Note: If "\( + c\)" is omitted, award no further marks.
evidence of substituting (2,0) into indefinite integral (M1)
e.g. \(2 = 0 + \sin 0 + c\) , \(c = 2\)
\(v(t) = {t^2} + \sin t + 2\) A1 N3
[5 marks]
\(\int {({t^2} + \sin t + 2)} {\rm{d}}t = \frac{{{t^3}}}{3} - \cos t + 2t\) A1A1A1
Note: Award A1 for each correct term.
evidence of using \(v(3) - v(0)\) (M1)
correct substitution A1
e.g. \((9 - \cos 3 + 6) - (0 - \cos 0 + 0)\) , \((15 - \cos 3) - ( - 1)\)
\(16 - \cos 3\) (accept \(p = 16\) , \(q = - 1\) ) A1A1 N3
[7 marks]
reference to motion, reference to first 3 seconds R1R1 N2
e.g. displacement in 3 seconds, distance travelled in 3 seconds
[2 marks]
Examiners report
Parts (a) and (b) of this question were generally well done.
Parts (a) and (b) of this question were generally well done.
Problems arose in part (c) with many candidates not substituting \(s(3) - s(0)\) correctly, leading to only a partially correct final answer. There were also a notable few who were not aware that \(\cos 0 = 1\) in both parts (a) and (c).
There were a variety of interesting answers about the motion of the particle, few being able to give both parts of the answer correctly.