## Fundamentals of Physics Extended (10th Edition)

$distance = 1.7\times 10^{2}$ $m$
They are travelling with an initial velocity of $37 m/s$ and this time, slowing at a rate of $4 m/s^{2}$. Let $d$ be the distance covered while slowing. We know the equation of kinematics, $v^{2}=v_{o}^{2}+2ad$ Putting the known values and solving, we get: $0^{2}=37^{2}+2(-4)d$ $\frac{-37^{2}}{-8}= d$ $d= 171.1$ $m$ $d \approx 1.7\times10^{2}$ $m$