Answer: \(0\text{ ms}^{-2}\), \(-3\text{ ms}^{-2}\), \(1650\text{ m}\); speed \(2\text{ ms}^{-1}\), \(t=\frac{\pi}{3}\), \(s=\frac43\sin3t-4t\).

(a)(i) At \(t=12\), the graph is horizontal because the particle is moving at constant velocity. Therefore the acceleration is

\(0\text{ ms}^{-2}.\)

(ii) From \(t=35\) to \(t=55\), the velocity decreases from \(60\) to \(0\). The acceleration is the gradient of the velocity-time graph:

\(\frac{0-60}{55-35}=-3.\)

So the acceleration when \(t=50\) is

\(-3\text{ ms}^{-2}.\)

(iii) The total distance is the area under the velocity-time graph.

From \(0\) to \(10\), the area is

\(\frac12(10)(30)=150.\)

From \(10\) to \(25\), the area is

\(15(30)=450.\)

From \(25\) to \(35\), the area is

\(\frac12(30+60)(10)=450.\)

From \(35\) to \(55\), the area is

\(\frac12(20)(60)=600.\)

Therefore the total distance is

\(150+450+450+600=1650\text{ m}.\)

(b)(i) The velocity is \(v=4\cos3t-4\). When \(t=\frac{5\pi}{9}\),

\(3t=\frac{5\pi}{3}.\)

So

The speed is the magnitude of velocity, so the speed is

\(2\text{ ms}^{-1}.\)

(ii) Acceleration is

\(a=\frac{dv}{dt}=-12\sin3t.\)

For the acceleration to be zero,

\(\sin3t=0.\)

The smallest positive value occurs when

\(3t=\pi.\)

Therefore

\(t=\frac{\pi}{3}.\)

(iii) Displacement is the integral of velocity:

\(s=\int(4\cos3t-4)\,dt.\)

Hence

\(s=\frac43\sin3t-4t+C.\)

Since \(Q\) passes through \(O\) when \(t=0\), \(s=0\) at \(t=0\). This gives \(C=0\). Therefore

\(s=\frac43\sin3t-4t.\)