Answer
$a=-1.3\,\mathrm{m/s^2}$
Work Step by Step
For the motion of the car on the dry pavement:
$\Delta x=50\,\mathrm{m}$, $a=-7.5\,\mathrm{m/s^2}$ and $v_0=70\,\mathrm{mi/h}=31\,\mathrm{m/s}$
We can find the final velocity of the car, when it reaches the junction between the dry pavement and the icy patch, using the kinematic equation:
\begin{align*}
v^2&=v_0^2+2a\Delta x\\
&=(31)^2+2(-7.5)(50)\\
&=961-750\\
&=211\\
v&=\sqrt{211}\\
v&=14.5\,\mathrm{m/s}
\end{align*}
This will be the initial speed of the car when it enters the icy patch, with $v=0$ (since it comes to rest) and $\Delta x=80\,\mathrm{m}$. Using the kinematic equation:
\begin{align*}
v^2&=v_0^2+2a\Delta x\\
0&=(14.5)^2+2a(80)\\
160a&=-210\\
a&=-{210\over 160}\\
a&=-1.3\,\mathrm{m/s^2}
\end{align*}
Deceleration on the ice is $1.3\,\mathrm{m/s^2}$
