a) Substitute \( t = 2 \) into \( v(t) = 1 + e^{-t} – e^{-\sin(2t)} \):
\( v(2) = 1 + e^{-2} – e^{-\sin(4)} \) [M1]
\( e^{-2} \approx 0.135335 \), \( \sin(4) \approx -0.756802 \), \( e^{-\sin(4)} \approx e^{0.756802} \approx 2.13147 \) [A1]
\( v(2) \approx 1 + 0.135335 – 2.13147 \approx -0.996135 \) [A1]
\( v = -0.996 \, \text{ms}^{-1} \) [A1] [3]

b) Differentiate \( v(t) \):
\( v'(t) = -e^{-t} + 2e^{-\sin(2t)}\cos(2t) \) [M1]
Set \( v'(t) = 0 \) to find critical points [M1]

Solve numerically: \( t \approx 0.405833 \) [A1]

Evaluate \( v(0.405833) \approx 1.18230 \) [A1]
Compare with endpoints: \( v(0) = 1 \), \( v(2) \approx -0.996135 \), confirming maximum at \( t \approx 0.405833 \) [M1]
\( v_{\text{max}} = 1.18 \, \text{ms}^{-1} \) [A1] [6]

c) Particle changes direction when \( v(t) = 0 \):
Solve \( 1 + e^{-t} – e^{-\sin(2t)} = 0 \), numerically \( t \approx 1.65840 \) [M1]
Acceleration: \( a(t) = v'(t) = -e^{-t} + 2e^{-\sin(2t)}\cos(2t) \) [M1]
Evaluate at \( t \approx 1.65840 \): \( a(1.65840) \approx -2.53487 \) [A1]
\( a = -2.53 \, \text{ms}^{-2} \) [A1] [4]