(i) To find \(k\), consider the velocity at \(t = 4\) from the first equation: \(v = 5t(t - 2)\). Substituting \(t = 4\), we get:

\(v = 5 \times 4 \times (4 - 2) = 40\).

Thus, \(k = 40\).

(ii) The velocity-time graph consists of three segments:

• For \(0 \leq t \leq 4\), the graph is a quadratic curve \(v = 5t(t - 2)\) with a minimum at \(t = 1\) and \(v = 0\) at \(t = 2\).
• For \(4 \leq t \leq 14\), the graph is a horizontal line at \(v = 40\).
• For \(14 \leq t \leq 20\), the graph is a line with a negative gradient \(v = 68 - 2t\).

(iii) For \(0 \leq t \leq 4\), the acceleration \(a = \frac{dv}{dt} = \frac{d}{dt}(5t(t - 2)) = 10t - 10\).

The acceleration is positive when \(10t - 10 > 0\), which gives \(t > 1\).

Thus, the set of values of \(t\) for which the acceleration is positive is \(1 < t < 4\).

(iv) To find the total distance travelled:

For \(0 \leq t \leq 4\), integrate \(v = 5t(t - 2)\):

\(\int (5t^2 - 10t) \, dt = \left[ \frac{5}{3}t^3 - 5t^2 \right]_0^4 = \frac{100}{3}\).

For \(4 \leq t \leq 14\), the distance is \(40 \times 10 = 400\).

For \(14 \leq t \leq 20\), integrate \(v = 68 - 2t\):

\(\int (68 - 2t) \, dt = \left[ 68t - t^2 \right]_{14}^{20} = 204\).

Total distance = \(\frac{100}{3} + 400 + 204 = 644\) m.