Explanation

(c) Assume that the motion is along the positive direction of $x$-axis. For simplicity, let us take the beginning of the braking to be a t time $t = 0,\ $at position $x_{0}$
Therefore, $x - x_{0} = v_{0}\left( \frac{v - v_{0}}{a} \right) + \frac{1}{2}a\left( \frac{v - v_{0}}{a} \right)^{2}$
Solving for $a$ and substituting known data then yield
$a = \frac{v^{2} - v_{0}}{2(x - x_{0})}$
Here, $v_{0} = 100kmh^{- 1} = 27.78ms^{- 1},x - x_{0} = 88.0$
And $v = 80kmh^{- 1} = 22.22m^{- 1}$
$\therefore\ a = \frac{(22.22)^{2} - (27.78)^{2}}{2(88.0)}$
$= - 1.58ms^{- 2}$