(a) [2 marks]

Particle changes direction when \( v(t) = 0 \) (M1).
\( v(t) = 2\sin(0.5t) + 0.3t – 2 = 0 \).
Numerically, \( t \approx 1.68694 \approx 1.69 \) seconds (A1). Acceleration \( a(t) = \cos(0.5t) + 0.3 > 0 \) at \( t \approx 1.69 \), confirming direction change.

(b) [2 marks]

Displacement increases when \( v(t) > 0 \) (M1).
From graph and solving \( v(t) = 0 \), zeros at \( t \approx 1.68694 \) and \( t \approx 6.11857 \).
\( v(t) > 0 \) for \( 1.68694 < t < 6.11857 \approx 1.69 < t < 6.12 \) (A1).

(c) [2 marks]

Displacement \( s(t) = \int_0^{10} v(t) \, dt = \int_0^{10} (2\sin(0.5t) + 0.3t – 2) \, dt \) (M1).
\( s(t) = -4\cos(0.5t) + 0.15t^2 – 2t + 4 \) (using \( s(0) = 0 \)).
At \( t = 10 \): \( s(10) \approx -4\cos(5) + 0.15 \cdot 100 – 20 + 4 \approx -2.13464 \approx -2.13 \, \text{m} \) (A1).