Find the displacement of the particle when \(t = 4\).

Sketch a displacement/time graph for the particle, \(0 \le t \le 5\), showing clearly where the curve meets the axes and the coordinates of the points where the displacement takes greatest and least values.

For \(t > 5\), the displacement of the particle is given by \(s = a + b\cos \frac{{2\pi t}}{5}\) such that \(s\) is continuous for all \(t \ge 0\).

Given further that \(s = 16.5\) when \(t = 7.5\), find the values of \(a\) and \(b\).

For \(t > 5\), the displacement of the particle is given by \(s = a + b\cos \frac{{2\pi t}}{5}\) such that \(s\) is continuous for all \(t \ge 0\).

Find the times \({t_1}\) and \({t_2}(0 < {t_1} < {t_2} < 8)\) when the particle returns to its starting point.

METHOD 1

\(s = \int {(9t - 3{t^2}){\text{d}}t = \frac{9}{2}{t^2} - {t^3}( + c)} \)     (M1)

\(t = 0,{\text{ }}s = 3 \Rightarrow c = 3\)     (A1)

\(t = 4 \Rightarrow s = 11\)     A1

METHOD 2

\(s = 3 + \int_0^4 {(9t - 3{t^2}){\text{d}}t} \)     (M1)(A1)

\(s = 11\)     A1

[3 marks]

correct shape over correct domain     A1

maximum at \((3,{\text{ }}16.5)\)     A1

\(t\) intercept at \(4.64\), \(s\) intercept at \(3\)     A1

minimum at \((5,{\text{ }} - 9.5)\)     A1

[5 marks]

\( - 9.5 = a + b\cos 2\pi \)

\(16.5 = a + b\cos 3\pi \)     (M1)

Note:     Only award M1 if two simultaneous equations are formed over the correct domain.

 \(a = \frac{7}{2}\)     A1

\(b =  - 13\)     A1

[3 marks]

at \({t_1}\):

\(3 + \frac{9}{2}{t^2} - {t^3} = 3\)     (M1)

\({t^2}\left( {\frac{9}{2} - t} \right) = 0\)

\({t_1} = \frac{9}{2}\)     A1

solving \(\frac{7}{2} - 13\cos \frac{{2\pi t}}{5} = 3\)     (M1)

\({\text{GDC}} \Rightarrow {t_2} = 6.22\)     A1

Note:     Accept graphical approaches.

[4 marks]

Total [15 marks]