## University Physics with Modern Physics (14th Edition)

$v = (105~km/h)(\frac{1000~m}{1~km})(\frac{1~h}{3600~s}) = 29.2~m/s$ $x = \frac{v^2-v_0^2}{2a} = \frac{0-(29.2~m/s)^2}{(2)(-250~m/s^2)}$ $x = 1.71~m$ The stopping distance must be at least 1.71 meters.