## Fundamentals of Physics Extended (10th Edition)

$distance = 3.4\times10^{2}$ $m$
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: $0^{2}=37^{2}+2(-2)d$ $\frac{-37^{2}}{-4} = d$ $d =342.25$ $m$ $\approx 3.4\times10^{2}$ $metres$