Can we find the velocity and acceleration of this particle?

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The velocity of the particle is the rate of change of its displacement from \(O\). So we can differentiate the given function with respect to time. \[\begin{align*} v&=\frac{ds}{dt}=\frac{3}{2}\times12\cos\left(\frac{3}{2}t\right) \\ &=18\cos\left(\frac{3}{2}t\right) \end{align*}\] We can differentiate this function now to give us an expression for the acceleration. \[\begin{align*} a&=\frac{dv}{dt}=-\frac{3}{2}\times18\sin\left(\frac{3}{2}t\right) \\ &=-27\sin\left(\frac{3}{2}t\right) \end{align*}\]

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

We can substitute this value into the given equation giving us \[\begin{align*} 6&=12\sin\left(\frac{3}{2}t\right) \\ \implies\quad \sin\left(\frac{3}{2}t\right)&=\frac{1}{2} \end{align*}\]

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 \[v=18\cos\left(\frac{3}{2}t\right)=18\frac{\sqrt{3}}{2}=9\sqrt{3}\approx\quantity{15.6}{m\,s^{-1}}.\]