(a)
\( a = \pi \approx 3.14 \)

Rest when: \( v = 3\sin(t)(1 + \cos(t)) = 0 \)
Solve: \( \sin(t) = 0 \) or \( 1 + \cos(t) = 0 \)
\( \sin(t) = 0 \): \( t = 0, \pi, 2\pi, \ldots \)
\( 1 + \cos(t) = 0 \): \( \cos(t) = -1 \), \( t = \pi, 3\pi, \ldots \)
Since \( t = 0 \) is start, next rest is \( t = \pi \)
Verify: At \( t = \pi \), \( v = 3 \times 0 \times (1 – 1) = 0 \)

Result: \( a = \pi \approx 3.14 \) [3]

(b)
Maximum velocity = \( \frac{9\sqrt{3}}{4} \approx 3.90 \, \text{ms}^{-1} \)

Velocity: \( v = 3\sin(t)(1 + \cos(t)) \), interval \( 0 \leq t \leq \pi \)
Derivative: \( \frac{dv}{dt} = 3 \left[ \cos(t)(1 + \cos(t)) – \sin(t)\sin(t) \right] \)
Simplify: \( = 3 \left[ \cos(t) + \cos^2(t) – \sin^2(t) \right] = 3 \left[ \cos(t) + \cos(2t) \right] \)
Set \( \frac{dv}{dt} = 0 \): \( \cos(t) + \cos(2t) = 0 \)
Use: \( \cos(2t) = 2\cos^2(t) – 1 \)
Solve: \( 2\cos^2(t) + \cos(t) – 1 = 0 \)
Let \( u = \cos(t) \): \( 2u^2 + u – 1 = 0 \)
Solve: \( u = \frac{1}{2} \) (\( t = \frac{\pi}{3} \)) or \( u = -1 \) (\( t = \pi \))
Evaluate: At \( t = 0 \), \( v = 0 \)
At \( t = \frac{\pi}{3} \): \( v = 3 \times \frac{\sqrt{3}}{2} \times \frac{3}{2} = \frac{9\sqrt{3}}{4} \approx 3.897 \)
At \( t = \pi \): \( v = 0 \)

Result: Maximum velocity = \( \frac{9\sqrt{3}}{4} \approx 3.90 \, \text{ms}^{-1} \) [3]

(c)
Average speed = \( \frac{6}{\pi} \approx 1.91 \, \text{ms}^{-1} \)

Distance: \( \int_0^\pi 3\sin(t)(1 + \cos(t)) \, dt \)
Since \( v \geq 0 \) in \( [0, \pi] \), use \( v \)
Substitute: \( u = 1 + \cos(t) \), \( du = -\sin(t) \, dt \)
Limits: \( t = 0 \), \( u = 2 \); \( t = \pi \), \( u = 0 \)
Integral: \( \int_0^2 3u \, du = 3 \times \frac{u^2}{2} \big|_0^2 = 3 \times 2 = 6 \)
Average speed: \( \frac{6}{\pi} \approx 1.90985 \)

Result: Average speed = \( \frac{6}{\pi} \approx 1.91 \, \text{ms}^{-1} \) [3]