(i) The acceleration is given by \(a(t) = 0.6t\). Integrating to find velocity:
\(v(t) = \int 0.6t \, dt = 0.3t^2 + C\).
Since the particle starts from rest, \(v(0) = 0\), so \(C = 0\). Thus, \(v(t) = 0.3t^2\).
At \(t = 10\), \(v(10) = 0.3 \times 10^2 = 30\) m s−1.
Integrating velocity to find displacement:
\(s(t) = \int 0.3t^2 \, dt = 0.1t^3 + C\).
Since \(s(0) = 0\), \(C = 0\). Thus, \(s(t) = 0.1t^3\).
At \(t = 10\), \(s(10) = 0.1 \times 10^3 = 100\) m.
(ii) After \(t = 10\), the acceleration is \(a(t) = -0.4t\). Integrating to find velocity:
\(v(t) = \int -0.4t \, dt = -0.2t^2 + C\).
Using \(v(10) = 30\), \(-0.2 \times 10^2 + C = 30\) gives \(C = 50\). Thus, \(v(t) = -0.2t^2 + 50\).
At point \(A\), \(v(t) = 0\):
\(-0.2t^2 + 50 = 0\) gives \(t = \sqrt{250}\).
Integrating velocity to find displacement:
\(s(t) = \int (-0.2t^2 + 50) \, dt = -\frac{t^3}{15} + 50t + C\).
Using \(s(10) = 100\), \(-\frac{10^3}{15} + 500 + C = 100\) gives \(C = -\frac{1000}{3}\).
Thus, \(s(t) = -\frac{t^3}{15} + 50t - \frac{1000}{3}\).
At \(t = \sqrt{250}\), \(s(\sqrt{250}) = -\frac{(\sqrt{250})^3}{15} + 50\sqrt{250} - \frac{1000}{3} = 194\) m.
