A particle sits on a smooth surface and is acted upon by a time dependent horizontal force, giving it an acceleration of a = 2t
2 + 4t where t is in seconds. Given that it is initially at rest and experiences no resistance
to motion, find:
a) The velocity of the particle at time t.
b) The distance travelled by the particle if acted on by the force for 8s

(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

no idea

step-by-step explanation:

answer: sherk

step-by-step explanation:

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