Answer
$$
s=54.0 \mathrm{~m}
$$
Work Step by Step
Velocity: $v_0=27 \mathrm{~m} / \mathrm{s}$ at $t_0=0 \mathrm{~s}$.
$$
$$
$$
\begin{gathered}
d v=a d t \\
\int_{27}^v d v=\int_0^t-6 t d t \\
v=\left(27-3 t^2\right) \mathrm{m} / \mathrm{s} &&(1)
\end{gathered}
$$
At $v=0$, from Eq. (1)
$$
0=27-3 t^2 \quad t=3.00 \mathrm{~s}
$$
Distance Traveled: $s_0=0 \mathrm{~m}$ at $t_0=0 \mathrm{~s}$. Using the result $v=27-3 t^2$ and applying Eq. 12-1, we have
$$
(+\downarrow)
$$
$$
\begin{aligned}
d s & =v d t \\
\int_0^s d s & =\int_0^t\left(27-3 t^2\right) d t \\
s & =\left(27 t-t^3\right) \mathrm{m}&(2)
\end{aligned}
$$
At $t=3.00 \mathrm{~s}$, from Eq. (2)
$$
s=27(3.00)-3.00^3=54.0 \mathrm{~m}
$$
