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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

User SorryMike
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1 Answer

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(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

User Matina
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