(i) For \(15 \leq t \leq 35\), the velocity is \(v = a + bt^2\). At \(t = 35\), \(v = 0\), so:

\(0 = a + b \times 35^2\)

\(0 = a + 1225b\)

At \(t = 15\), \(v = 2 \times 15 + 10 = 40\), so:

\(40 = a + b \times 15^2\)

\(40 = a + 225b\)

Solving these equations:

\(1000b = -40 \rightarrow b = -0.04\)

\(a = 0.04 \times 352 = 49\)

(ii) The velocity-time graph consists of:

• \(0 \leq t \leq 5\): Increasing quadratic from (0,0) to (5,20), concave up.
• \(5 \leq t \leq 15\): Line from (5,20) to (15,40).
• \(15 \leq t \leq 35\): Decreasing quadratic from (15,40) to (35,0), concave down.

(iii) Total distance:

\(A_1 = \int_0^5 0.8t^2 \, dt = \frac{100}{3}\)

\(A_2 = \frac{1}{2} (20 + 40) \times 10 = 300\)

\(A_3 = \int_{15}^{35} (a + bt^2) \, dt = 49t - \frac{0.04}{3}t^3 \bigg|_{15}^{35}\)

\(A_3 = 453.3333 = \frac{1360}{3}\)

Total Distance = \(\frac{100}{3} + 300 + \frac{1360}{3} = 787 \text{ m}\)