A particle moves along a line so that its acceleration for \(t\geq 0\) is given by \(a(t)=\dfrac{t+2}{\sqrt{t^3+3}} \). If the particle's velocity at \(t=0\) is \(10\), what is the velocity of the particle at \(t=5\)?

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The change in velocity can be found using:

$$ \int\limits_0^5 a(t)   dt $$

The final velocity is:

$$ 10 + \int\limits_0^5 a(t)=\dfrac{t+2}{\sqrt{t^3+3}}   dt $$ $$ \approx 10 + 5.296 $$ $$ = \boxed{10.296} $$