The distance traveled during a period of constant acceleration is given by
[tex]d=\dfrac{(v_{2})^{2}-(v_{1})^{2}}{2a}[/tex]
Your initial speed is 50 mi/h = 73 1/3 ft/s, so the stopping distance can be computed as
[tex]d=\dfrac{0-(73\frac{1}{3})^{2}}{2\cdot(-26)}=\dfrac{5777\frac{7}{9}}{52}\approx 103.4[/tex]
The stopping distance is 103.4 ft.
