Answer
The stopping distance is $~~100~m$
Work Step by Step
We can express the initial velocity in units of $m/s$:
$v_0 = (113~km/h)(\frac{1~h}{3600~s})(\frac{1000~m}{1~km}) = 31.4~m/s$
We can find the magnitude of deceleration:
$F = ma$
$a = \frac{F}{m}$
$a = \frac{F}{weight/g}$
$a = \frac{F~g}{weight}$
$a = \frac{(8230~N)(9.8~m/s^2)}{16,400~N}$
$a = 4.92~m/s^2$
We can find the stopping distance:
$v_f^2 = v_0^2+2ad$
$0 = v_0^2+2ad$
$-2ad = v_0^2$
$d = \frac{v_0^2}{-2a}$
$d = \frac{(31.4~m/s)^2}{-(2)(-4.92~m/s^2)}$
$d = 100~m$
The stopping distance is $~~100~m$
