a) To find the smallest \( t \) when the particle changes direction:
A direction change occurs when velocity \( v = 0 \) and the sign of velocity changes, confirmed by acceleration.
Velocity: \( v = 2\sin(0.5t) + 0.3t – 2 \)
Set \( v = 0 \):
\( 2\sin(0.5t) + 0.3t – 2 = 0 \)

The graph shows a zero crossing between \( t = 1 \) and \( t = 2 \). Test values:
At \( t = 1 \):
\( v = 2\sin(0.5 \times 1) + 0.3 \times 1 – 2 \approx 2 \times 0.4794 + 0.3 – 2 \approx 0.9588 – 1.7 \approx -0.7412 \) (negative)
At \( t = 2 \):
\( v = 2\sin(0.5 \times 2) + 0.3 \times 2 – 2 \approx 2 \times 0.8415 + 0.6 – 2 \approx 1.683 + 0.6 – 2 \approx 0.283 \) (positive)
At \( t = 1.68694 \):
\( v = 2\sin(0.5 \times 1.68694) + 0.3 \times 1.68694 – 2 \approx 2 \times 0.7465 + 0.506082 – 2 \approx 0 \)

Acceleration: \( a = \frac{dv}{dt} = \cos(0.5t) + 0.3 \)
At \( t = 1.68694 \):
\( a \approx \cos(0.5 \times 1.68694) + 0.3 \approx 0.6658 + 0.3 \approx 0.9658 \) (positive)

Velocity changes from negative to positive at \( t \approx 1.68694 \), confirmed by positive acceleration.
Smallest \( t \):
\( t \approx 1.69 \) seconds (to two decimal places) [3]

b) To find when displacement is increasing:
Displacement increases when velocity \( v > 0 \):
\( v = 2\sin(0.5t) + 0.3t – 2 \)

From part (a), first root at \( t \approx 1.68694 \). Find the second root using the graph:
At \( t = 6 \):
\( v \approx 2 \times 0.1411 + 1.8 – 2 \approx 0.0822 \) (positive)
At \( t = 6.5 \):
\( v \approx 2 \times (-0.1083) + 1.95 – 2 \approx -0.2666 \) (negative)
At \( t = 6.11857 \):
\( v \approx 2 \times 0.0814 + 0.3 \times 6.11857 – 2 \approx 0 \)

Velocity is positive between roots:
\( 1.68694 < t < 6.11857 \)
Rounded to two decimal places:
\( 1.69 < t < 6.12 \) [3]

c) To find displacement at \( t = 10 \):
Displacement \( s(t) = \int v(t) \, dt \), with \( s(0) = 0 \):
\( v = 2\sin(0.5t) + 0.3t – 2 \)

Integrate:
\( \int 2\sin(0.5t) \, dt = -4\cos(0.5t) \)
\( \int 0.3t \, dt = 0.15t^2 \)
\( \int -2 \, dt = -2t \)
\( s(t) = -4\cos(0.5t) + 0.15t^2 – 2t + C \)

At \( t = 0 \), \( s(0) = 0 \):
\( -4\cos(0) + 0 – 0 + C = -4 + C = 0 \)
\( C = 4 \)
\( s(t) = -4\cos(0.5t) + 0.15t^2 – 2t + 4 \)

At \( t = 10 \):
\( s(10) = -4\cos(0.5 \times 10) + 0.15 \times 10^2 – 2 \times 10 + 4 \)
\( = -4\cos(5) + 15 – 20 + 4 \)
\( \cos(5) \approx 0.2837 \)
\( s(10) \approx -4 \times 0.2837 – 1 \approx -1.1348 – 1 \approx -2.1348 \)

Displacement:
\( s(10) \approx -2.13 \) meters (to two decimal places) [3]