(a) By the fundamental theorem of calculus,
v(t) = v(0) + ∫₀ᵗ a(u) du
The particle starts at rest, so v(0) = 0. Computing the integral gives
v(t) = [2/3 u ³ + 2u ²]₀ᵗ = 2/3 t ³ + 2t ²
(b) Use the FTC again, but this time you want the distance, which means you need to integrate the speed of the particle, i.e. the absolute value of v(t). Fortunately, for t ≥ 0, we have v(t) ≥ 0 and |v(t) | = v(t), so speed is governed by the same function. Taking the starting point to be the origin, after 8 seconds the particle travels a distance of
∫₀⁸ v(u) du = ∫₀⁸ (2/3 u ³ + 2u ²) du = [1/6 u ⁴ + 2/3 u ³]₀⁸ = 1024