Answer
The distance is 211.2 ft.
Work Step by Step
Let the acceleration be a, final velocity be v, initial velocity be u and displacement be s
Using the kinematics equations, $v^{2}=u^{2}+2as$
Convert 45 and 30 mph to ft/s
45 mph = 45$\times\frac{5280}{3600}$=66ft/s
30 mph = 30$\times\frac{5280}{3600}$=44ft/s
a=$\frac{(44)^{2}-(66)^{2}}{2(264)}$
a=-$\frac{55}{12}$
To find the stopping distance, set v=0, u=44
0=44$^{2}$+2(-$\frac{55}{12}$)s
s=$\frac{(44)^{2}}{2(\frac{55}{12})}$
=211.2 ft
