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]

[N/A]