(i) To find the deceleration, we first calculate the frictional force using the formula:

\(F = ext{coefficient of friction} imes ext{normal force} = 0.2 imes 6g imes ext{cos}(8^ ext{o})\)

Using Newton's second law, the net force along the plane is:

\(6g imes ext{sin}(8^ ext{o}) - F = 6a\)

Solving for acceleration \(a\), we find:

\(a = rac{6g imes ext{sin}(8^ ext{o}) - F}{6}\)

Substituting the values, we get:

\(a = 0.589 ext{ m s}^{-2}\)

(ii) To find the distance traveled, we use the equation of motion:

\(v^2 = u^2 + 2as\)

Given that the final velocity \(v = 0\), initial velocity \(u = 3 ext{ m s}^{-1}\), and \(a = -0.589 ext{ m s}^{-2}\), we solve for \(s\):

\(0 = 3^2 + 2(-0.589)s\)

Solving for \(s\), we find:

\(s = 7.64 ext{ m}\)

(a) For the motion from A to B, the potential energy (PE) loss is converted into kinetic energy (KE):

\(0.2 \times 10 \times 0.5 = \frac{1}{2} \times 0.2 \times v_B^2\)

Solving gives \(v_B^2 = 10\).

For the motion from B to C, the frictional force \(F\) is given by:

\(R = 0.2 \times 10 \times \cos 30 = \sqrt{3}\)

\(F = \mu R = \frac{\sqrt{3}}{2} \times \sqrt{3} = 1.5\)

The work done against friction is \(1.5 \times 1\).

Using the work-energy principle:

\(\frac{1}{2} \times 0.2 \times 10 + 0.2 \times 10 \times 0.5 = 1.5 \times 1 + \frac{1}{2} \times 0.2 \times v_c^2\)

Solving gives \(v_c = \sqrt{5} \text{ m/s}\).

(b) If the particle comes to rest at C, then:

\(0 = 10 + 2a\)

\(a = -5\)

Using Newton's second law:

\(0.2 \times 10 \times \sin 30 - F = 0.2 \times -5\)

\(2 = \mu \sqrt{3}\)

Solving gives \(\mu = \frac{2}{\sqrt{3}}\).

(i) Using the equation of motion \(s = ut + \frac{1}{2}at^2\) for the segment PQ:

\(0.8 = 0.6u + \frac{1}{2}a(0.6)^2\)

\(0.8 = 0.6u + 0.18a\)

For the segment QR:

\(0.8 = (u + a \times 0.6) \times 1 + \frac{1}{2}a \times 1^2\)

\(0.8 = 1.6u + 1.28a\)

Solving these equations simultaneously gives:

Deceleration \(a = \frac{2}{3}\) m s^-2

\(u = \frac{23}{15}\) m s^-1

(ii) The normal reaction \(R = mg \cos 3\)

The frictional force \(F = \mu mg \cos 3\)

Using Newton's second law:

\(-mg \sin 3 - \mu mg \cos 3 = m \left( -\frac{2}{3} \right)\)

Solving for \(\mu\):

\(\mu = 0.0144\)

(i) To find the acceleration, we start by calculating the normal reaction force, which is given by:

\(R = mg \cos 25^{\circ}\)

The frictional force \(F\) is given by:

\(F = 0.4mg \cos 25^{\circ}\)

Using Newton's Second Law along the plane:

\(mg \sin 25^{\circ} - 0.4mg \cos 25^{\circ} = ma\)

Solving for \(a\):

\(a = g(\sin 25^{\circ} - 0.4 \cos 25^{\circ})\)

Substituting \(g = 9.8 \text{ m/s}^2\), we find:

\(a = 0.601 \text{ m/s}^2\)

(ii) To find the distance travelled in the first 3 seconds, we use the equation:

\(s = ut + \frac{1}{2}at^2\)

Since the initial velocity \(u = 0\), the equation simplifies to:

\(s = \frac{1}{2} \times 0.601 \times 3^2\)

Calculating this gives:

\(s = 2.70 \text{ m}\)

(i) Using the equation of motion, \(v = u + at\), where \(v = 2 \text{ m/s}\), \(u = 0 \text{ m/s}\), and \(t = 5 \text{ s}\), we have:

\(2 = 0 + 5a\)

\(a = 0.4 \text{ m/s}^2\)

Applying Newton's second law along the plane:

\(0.1g \sin 20^{\circ} - F = 0.1 \times 0.4\)

\(F = 0.1g \sin 20^{\circ} - 0.04\)

\(F = 0.302 \text{ N}\)

(ii) The normal reaction \(R\) is given by:

\(R = 0.1g \cos 20^{\circ} \approx 0.9397 \text{ N}\)

The coefficient of friction \(\mu\) is:

\(\mu = \frac{F}{R} = \frac{0.302}{0.9397} \approx 0.321\)