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

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

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:

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