Chapter 2 - One-Dimensional Kinematics - Problems and Conceptual Exercises - Page 53: 71

Since $2.8~s\approx3~s$, the statment is accurate.

$v_0=0$ and $a=g=9.81~m/s^2$ $1~mi=1609~m$ and $1~h=3600~s$ $60~mi/h=(60~mi/h)(\frac{1609~m}{1~mi})(\frac{1~h}{3600~s})=27~m/s$ $v=v_0+at=gt$ $27~m/s=(9.81~m/s^2)t$ $t=\frac{27~m/s}{9.81~m/s^2}=2.8~s$ Since $2.8~s\approx3~s$, the statment is accurate.