Chapter 1 - Section 1.12 - Modeling Variation - 1.12 Exercises - Page 128: 39

$$\approx46.19mi/h$$

According to the problem, we have: $D=ks^2$ $D$ is stopping distance $s$ is speed $k$ is constant of proportionality --- $s=40mi/h$ $D=150ft$ $k$ is unknown $150=k\times 40^2$ $1600k=150$ $k=0.09375 ft\times mi^2/h^2$ So, we have a general formula $D=0.09375\times s^2$ If $D=200ft$ $200=0.09375\times s^2$ $s^2=2133.(3)$ $s\approx46.19mi/h$