To find T₁ and T₂, we first find the acceleration function by differentiating the velocity function:

\(a(t) = \frac{d}{dt}(0.002t^3 - 0.12t^2 + 1.8t + 5) = 0.006t^2 - 0.24t + 1.8\)

Set \(a(t) = 0\) to find critical points:

\(0.006t^2 - 0.24t + 1.8 = 0\)

Solving the quadratic equation:

\(0.006(t^2 - 40t + 300) = 0\)

\(t^2 - 40t + 300 = 0\)

Using the quadratic formula:

\(t = \frac{40 \pm \sqrt{40^2 - 4 \times 1 \times 300}}{2 \times 1}\)

\(t = \frac{40 \pm \sqrt{1600 - 1200}}{2}\)

\(t = \frac{40 \pm 20}{2}\)

\(t = 30\) or \(t = 10\)

Thus, \(T_1 = 10\) and \(T_2 = 30\).

To find the distance \(OP\) when \(t = T_2\), integrate the velocity function:

\(s(t) = \int (0.002t^3 - 0.12t^2 + 1.8t + 5) \, dt\)

\(s(t) = 0.0005t^4 - 0.04t^3 + 0.9t^2 + 5t + C\)

Using initial condition \(s(0) = 0\), \(C = 0\).

Calculate \(s(T_2)\):

\(s(30) = 0.0005(30)^4 - 0.04(30)^3 + 0.9(30)^2 + 5(30)\)

\(s(30) = 405 - 1080 + 810 + 150\)

\(s(30) = 285\) m

For part (ii), the velocity at \(t = T_2\) is:

\(v(30) = 0.002(30)^3 - 0.12(30)^2 + 1.8(30) + 5\)

\(v(30) = 5\) m s^-1

The velocity-time graph shows \(v\) increasing from a positive value at \(t = 0\) to a maximum, then decreasing to a positive minimum, and thereafter increasing.