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Solution

Given

a(t) = 6t^{2} + 1

Initial Velocity (u) = 4m/sec

initial displacement (s(0)) = 5cm

Position function (s(t)) = ?

We know

v'(t) = a(t) = 6t^{2} + 1

Integrating we get,

\(v(t) = 6\frac{t^3}{3} + t + C\)

\(v(t) = 2t^3 + t + C\)

As we know, v(0) = 4m/sec then c = 4

\(v(t) = 2t^3 + t + 4\)

Again

s(t) = ∫v(t) dt

s(t) = ∫(2t^{3} + t + 4) dt

\(s(t) = \frac{t^4}{2} + \frac{t^2}{2} + 4t + D\)

Since s(0) = 5 so D = 5

Therefore position function is given by

\(s(t) = \frac{t^4}{2} + \frac{t^2}{2} + 4t + 5\)

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## Discussion