Note:     In this question, distance is in metres and time is in seconds.

A particle moves along a horizontal line starting at a fixed point A. The velocity $v$ of the particle, at time $t$, is given by $v(t) = \frac{{2{t^2} - 4t}}{{{t^2} - 2t + 2}}$, for $0 \leqslant t \leqslant 5$. The following diagram shows the graph of $v$

There are $t$-intercepts at $(0,{\text{ }}0)$ and $(2,{\text{ }}0)$.

Find the maximum distance of the particle from A during the time $0 \leqslant t \leqslant 5$ and justify your answer.

METHOD 1 (displacement)

recognizing $s = \int {v{\text{d}}t}$     (M1)

consideration of displacement at $t = 2$ and $t = 5$ (seen anywhere)     M1

eg$\,\,\,\,\,$$\int_0^2 v$ and $\int_0^5 v$

Note:     Must have both for any further marks.

correct displacement at $t = 2$ and $t = 5$ (seen anywhere)     A1A1

$- 2.28318$ (accept 2.28318), 1.55513

valid reasoning comparing correct displacements     R1

eg$\,\,\,\,\,$$\left| { - 2.28} \right| > \left| {1.56} \right|$, more left than right

2.28 (m)     A1     N1

Note:     Do not award the final A1 without the R1.

METHOD 2 (distance travelled)

recognizing distance $= \int {\left| v \right|{\text{d}}t}$     (M1)

consideration of distance travelled from $t = 0$ to 2 and $t = 2$ to 5 (seen anywhere)     M1

eg$\,\,\,\,\,$$\int_0^2 v$ and $\int_2^5 v$

Note:     Must have both for any further marks

correct distances travelled (seen anywhere)     A1A1

2.28318, (accept $- 2.28318$), 3.83832

valid reasoning comparing correct distance values     R1

eg$\,\,\,\,\,$$3.84 - 2.28 < 2.28,{\text{ }}3.84 < 2 \times 2.28$

2.28 (m)     A1     N1

Note:     Do not award the final A1 without the R1.

[6 marks]