\frametitle{Antiderivatives / Integrals}

    A particle moves in a straight line and has 
      \item acceleration $a(t) = 6t + 4$
      \item initial velocity is $v(0) = -6$cm/s
      \item initial displacement is $s(0) = 9$cm
    Find the position function $s(t)$.
    The velocity is an antiderivative of the acceleration:
      v(t) = \mpause[1]{3t^2} \mpause{+ 4t}\mpause{ + C}
    As $v(0) = -6$cm/s, it follows that $C = \pause -6$.
    The position function is an antiderivative of the velocity:
      s(t) = \mpause[1]{t^3} \mpause{+ 2t^2}\mpause{ - 6t}\mpause{+ D}
    As $s(0) = 9$cm/s, it follows that $D = \pause 9$.
    Thus the position function is:
      s(t) = t^3 + 2t^2 - 6t + 9 \quad \text{in cm}