A particle moves in a straight line. Its velocity \(v{\text{ m}}\,{{\text{s}}^{ - 1}}\) after \(t\) seconds is given by

\[v = 6t - 6,{\text{ for }}0 \leqslant t \leqslant 2.\]

After \(p\) seconds, the particle is 2 m from its initial position. Find the possible values of \(p\).

correct approach     (A1)

eg\(\,\,\,\,\,\)\(s = \int {v,{\text{ }}\int_0^p {6t - 6{\text{d}}t} } \)

correct integration     (A1)

eg\(\,\,\,\,\,\)\(\int {6t - 6{\text{d}}t = 3{t^2} - 6t + C,{\text{ }}\left[ {3{t^2} - 6t} \right]_0^p} \)

recognizing that there are two possibilities     (M1)

eg\(\,\,\,\,\,\)2 correct answers, \(s =  \pm 2,{\text{ }}c \pm 2\)

two correct equations in \(p\)     A1A1

eg\(\,\,\,\,\,\)\(3{p^2} - 6p = 2,{\text{ }}3{p^2} - 6p =  - 2\)

0.42265, 1.57735

\(p = 0.423{\text{ or }}p = 1.58\)    A1A1     N3

[7 marks]

Most candidates realized that they needed to calculate the integral of the velocity, and did it correctly. However, only a few realized that there were two possible positions for the particle, as it could move in two directions. In general, the only equation candidates wrote was \(3{p^2} - 6p = 2\), that gave solutions outside the given domain. Candidates failed to differentiate between displacement and distance travelled.