The function position of the particle is s(t) = (1 / 12) · t⁴ + (7 / 6) · t³ + 3 · t² + (63 / 4) · t.
By mechanical physics we know that the function velocity is the integral of function acceleration and the function position is the integral of function velocity. Hence, we need to integrate twice to obtain the function position of the particle:
Velocity
v(t) = ∫ t² dt - 7 ∫ t dt + 6 ∫ dt
v(t) = (1 / 3) · t³ - (7 / 2) · t² + 6 · t + C₁
Position
s(t) = (1 / 3) ∫ t³ dt - (7 / 2) ∫ t² dt + 6 ∫ t dt + C₁ ∫ dt
s(t) = (1 / 12) · t⁴ + (7 / 6) · t³ + 3 · t² + C₁ · t + C₂
Now we find the values of the integration constants by solving the following system of linear equations:
0 = C₂
63 / 4 = C₁ + C₂
The solution of the system is C₁ = 63 / 4 and C₂ = 0. The function position of the particle is s(t) = (1 / 12) · t⁴ + (7 / 6) · t³ + 3 · t² + (63 / 4) · t.
To learn more on parametric equations: https://brainly.com/question/9056657
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