(a) To find the value of \(t\) at point B, we need to determine when the velocity \(v = 0\). The velocity is the integral of acceleration:

\(v = \int (6t^{\frac{1}{2}} - 2t) \, dt\)

\(v = 4t^{\frac{3}{2}} - t^2 + c\)

Since the particle starts from rest, \(v = 0\) when \(t = 0\), so \(c = 0\).

At point B, \(v = 0\):

\(4t^{\frac{3}{2}} - t^2 = 0\)

\(t^{\frac{1}{2}}(4t^{\frac{1}{2}} - t) = 0\)

\(t^{\frac{1}{2}} = 0 \text{ or } 4t^{\frac{1}{2}} = t\)

\(t = 0 \text{ or } t = 16\)

Thus, \(t = 16\) at point B.

(b) To find the distance travelled from A to the point where the acceleration is zero again, set \(a = 0\):

\(6t^{\frac{1}{2}} - 2t = 0\)

\(t^{\frac{1}{2}}(6 - 2t^{\frac{1}{2}}) = 0\)

\(t^{\frac{1}{2}} = 0 \text{ or } 6 = 2t^{\frac{1}{2}}\)

\(t = 0 \text{ or } t = 9\)

Integrate the velocity to find the distance \(s\):

\(s = \int (4t^{\frac{3}{2}} - t^2) \, dt\)

\(s = \left[ \frac{8}{5}t^{\frac{5}{2}} - \frac{1}{3}t^3 \right]\)

Evaluate from \(t = 0\) to \(t = 9\):

\(s = \frac{8}{5}(9)^{\frac{5}{2}} - \frac{1}{3}(9)^3\)

\(s = 145.8 \text{ m}\)

(a) To find the displacement, integrate the velocity function:

\(s = \int (t^2 - 8t^{3/2} + 10t) \, dt\)

\(s = \left[ \frac{t^3}{3} - \frac{16}{5}t^{5/2} + 5t^2 \right] + C\)

Using the limits from \(t = 0\) to \(t = 1\),

\(s = \left( \frac{1^3}{3} - \frac{16}{5} \times 1^{5/2} + 5 \times 1^2 \right) - \left( \frac{0^3}{3} - \frac{16}{5} \times 0^{5/2} + 5 \times 0^2 \right)\)

\(s = \frac{1}{3} - \frac{16}{5} + 5 = \frac{32}{15} \approx 2.13 \text{ m}\)

(b) To find the minimum velocity, differentiate the velocity function:

\(a = \frac{dv}{dt} = 2t - 12t^{1/2} + 10\)

Set \(a = 0\):

\(2t - 12t^{1/2} + 10 = 0\)

Solving this quadratic equation in terms of \(\sqrt{t}\),

\(2(\sqrt{t} - 5)(\sqrt{t} - 1) = 0\)

\(t = 1\) or \(t = 25\)

Check the nature of the stationary point by differentiating \(a\):

\(\frac{da}{dt} = 2 - 6t^{-1/2}\)

At \(t = 25\), \(\frac{da}{dt} = 0.8\), indicating a minimum.

Evaluate \(v\) at \(t = 25\):

\(v = 25^2 - 8 \times 25^{3/2} + 10 \times 25 = -125 \text{ m s}^{-1}\)

\((a) To find when P is at instantaneous rest, set v = 0:\)

\(4t^2 - 20t + 21 = 0\)

Factorize the quadratic:

\((2t - 3)(2t - 7) = 0\)

\(Thus, t = 1.5 and t = 3.5.\)

(b) The acceleration a is given by the derivative of v:

\(a = \frac{dv}{dt} = 8t - 20\)

\(At t = 0, a = 8(0) - 20 = -20.\)

(c) To find the minimum velocity, set the derivative of v to zero:

\(8t - 20 = 0\)

Solve for t:

\(t = 2.5\)

Substitute back into v:

\(v = 4(2.5)^2 - 20(2.5) + 21 = -4\)

\(Thus, v_{min} = -4 \text{ ms}^{-1}.\)

(d) Integrate v to find the distance:

\(s = \int (4t^2 - 20t + 21) \, dt = \frac{4}{3}t^3 - 10t^2 + 21t + c\)

\(Evaluate from t = 1.5 to t = 3.5:\)

\(s = \left[ \frac{4}{3}(3.5)^3 - 10(3.5)^2 + 21(3.5) \right] - \left[ \frac{4}{3}(1.5)^3 - 10(1.5)^2 + 21(1.5) \right]\)

Calculate the distance:

\(s = \frac{49}{6} - \frac{27}{2} = \frac{16}{3} = 5.33 \text{ m}\)

(a) To find the velocity, integrate the acceleration:

\(v = \int 0.1t^{3/2} \, dt = 0.04t^{5/2} + 1.72\)

Set \(v = 3\) and solve for \(t\):

\(0.04t^{5/2} + 1.72 = 3\)

\(0.04t^{5/2} = 1.28\)

\(t^{5/2} = 32\)

\(t = 4\)

(b) To find the displacement, integrate the velocity:

\(s = \int (0.04t^{5/2} + 1.72) \, dt\)

\(s = \left[ \frac{2}{175}t^{7/2} + 1.72t \right]\)

Evaluate from \(t = 0\) to \(t = 2\):

\(s = \left[ \frac{2}{175}(2)^{7/2} + 1.72(2) \right]\)

\(s = 3.57 \text{ m}\)

Given the acceleration function: \(a = 6t - 18\).

To find the velocity \(v\), integrate the acceleration:

\(v = \int (6t - 18) \, dt = 3t^2 - 18t + C\).

Since the particle starts from rest, \(v = 0\) when \(t = 0\), so \(C = 0\).

Thus, \(v = 3t^2 - 18t\).

To find when the particle comes to rest, set \(v = 0\):

\(3t^2 - 18t = 0\).

Factor the equation: \(3t(t - 6) = 0\).

So, \(t = 0\) or \(t = 6\). The particle comes to rest at \(t = 6\) (ignoring \(t = 0\) as it is the starting point).

To find the distance \(s\), integrate the velocity:

\(s = \int (3t^2 - 18t) \, dt = t^3 - 9t^2 + C\).

Since \(s = 0\) when \(t = 0\), \(C = 0\).

Thus, \(s = t^3 - 9t^2\).

Substitute \(t = 6\) to find the distance:

\(s = 6^3 - 9 \times 6^2 = 216 - 324 = -108\).

Since distance must be positive, \(s = 108\) m.