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]