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.