Answer
$\bar{a}=480\,\mathrm{m/s^2}$
Work Step by Step
Initial velocity, $v_1=-35\,\mathrm{m/s}$
Final velocity, $v_2=11\,\mathrm{m/s}$
Change in velocity, $\Delta v=v_2-v_1=11-(-35)=46\,\mathrm{m/s}$
Time taken for the velocity change, $\Delta t=0.095\,\mathrm{s}$
Average acceleration
\begin{align*}
\bar{a}&={\Delta v\over\Delta t}\\
&={46\over 0.095}\\
\bar{a}&=480\,\mathrm{m/s^2}
\end{align*}
This value for average acceleration of the puck is much larger than the typical accelerations of cars for two reasons:
1. The puck is not just slowing down but reversing direction.
2. The change in velocity including the reversal of direction is taking too short a time--less than 1/10 of a second.
Even though cars travel at speeds similar to that of the puck, they do not generally reverse directions so quickly (except during collisions), and so their accelerations are not usually this large in magnitude.
