A particle \(P\) moves in a straight line so that, after \(t\) seconds, its distance, \(s\) metres, from a fixed point \(O\) is given by the equation. \[s=12\sin\left(\frac{3}{2}t\right)\] Determine

- expressions for the velocity and acceleration of \(P\) in terms of \(t\),

- the time at which \(P\) first returns to \(O\),

The graph of \(s\) is a sine wave that has been stretched horizontally by a factor of \(\frac{2}{3}\). A graph of \(\sin t\) would meet the axis at \(t=\pi\) so this graph will meet the axis at \[t=\pi\times\frac{2}{3}\] The particle first returns to \(0\) at \(t=\frac{2}{3}\pi\approx\quantity{2.1}{s}\).

- the maximum distance of \(P\) from \(O\) during the motion,

The graph has been stretched vertically so that it has an amplitude of \(12\). The maximum distance from \(O\) is equal to \(\quantity{12}{m}\).

- the velocity of \(P\) when \(s=6\).

Using this standard triangle we can see that if \(\sin\left(\frac{3}{2}t\right)=\frac{1}{2}\) then \(\cos\left(\frac{3}{2}t\right)=\frac{\sqrt{3}}{2}\). Therefore using the expression from part (1) we can calculate the velocity as \[v=18\cos\left(\frac{3}{2}t\right)=18\frac{\sqrt{3}}{2}=9\sqrt{3}\approx\quantity{15.6}{m\,s^{-1}}.\]