Originally Posted by
SlipEternal $a(t) = 3t$
Acceleration is the derivative of velocity: $a(t) = v'(t)$. So:
$\displaystyle v(t) = \int a(t) dt = \dfrac{3}{2}t^2+C$
You are told initial velocity is 5. Initial velocity means t=0. So:
$v(0) = 5 = \dfrac{3}{2}(0)^2 + C$
shows $C=5$.
Hence, $v(t) = \dfrac{3}{2}t^2+5$.
Velocity is the derivative of position: $v(t) = s'(t)$. So, $\displaystyle s(t) = \int v(t)dt$.
Now, you are looking for the distance the particle has moved. That is just $\displaystyle s(3)-s(0) = \int_0^3 v(t)dt$:
$\displaystyle \int_0^3 v(t)dt = \int_0^3 \left(\dfrac{3}{2}t^2+5\right)dt = \left[\dfrac{1}{2}t^3 + 5t\right]_0^3 = 19.5$