Answer
$distance = 3.4\times10^{2}$ $m$
Work Step by Step
They are travelling with an initial velocity of $37m/s $ and with a retardation of $2m/s^{2}$. Let $d$ be distance covered while slowing.
We know that,
$v^{2}=v_{o}^{2}+2a d$
We substitute the values in the formula above and solve:
$0^{2}=37^{2}+2(-2)d$
$\frac{-37^{2}}{-4} = d$
$d =342.25$ $m$ $\approx 3.4\times10^{2}$ $metres$
