Answer
s(t)= 279.67 m
Work Step by Step
By using the position and velocity particle functions:
\[
\begin{array}{c}
s(t)=\frac{1}{2} a t^{2} +s_{0}+v_{0} t\\
v(t)=a t+v_{0}
\end{array}
\]
We will find the time that the motorcycle take to travel $120 \mathrm{m}$
\[
s(t)=\frac{1}{2} a t^{2}+v_{0} t+s_{0}
\]
\[
\begin{aligned}
120 &=\frac{2.6}{2} t^{2}+0+0 \cdot t \\
&9.61=t
\end{aligned}
\]
$\therefore v(t)=a t+v_{0}=0+9.6 \cdot 2.6=24.96 \mathrm{m} / \mathrm{s}$
The motorcycle slows down with $v=12 m / s$, $a=1.5 \mathrm{m} / \mathrm{s}^{2}$
We have $s_{0}=120 \mathrm{m}, v_{0}=24.96 \mathrm{m} / \mathrm{s}$
$\therefore a t+v_{0}=v(t)$
$\Rightarrow -1.5 t+24.96=12$
$\Rightarrow t=[8.64]$
So, the distance traveled by the motorcycle is :
$s(t)=v_{0} t+\frac{1}{2} a t^{2}+s_{0}=-\frac{1.5}{2}(8.64)^{2}+24.96 \cdot 8.64+120=279.67$
