A ranger in a national park is driving at 11.8mi / h when a deer jumps into the road 242 ft ahead of the vehicle. After a reaction time of t the ranger applies the brakes to produce and acceleration of -9.18ft/s^2 What is the maximum reaction time allowed if she is to avoid hitting the deer? Answer in units of s.

In order to answer this problem let's start by converting the speed of the ranger's vehicle from miles per hour into feet per second, knowing that 1 mile is the same as 5280 ft and i hour is 3600 seconds:

[tex]11.8\, \frac{mi}{h} =11.8 \,\frac{5280\,ft}{3600\,s} \approx 17.3\,\frac{ft}{s}[/tex]

Now, with this information, we set the equation for the amount of time needed to reduce the speed from 17.3 ft/s to full stop (0 ft/s):

[tex]v_f-v_i=-9.18 \,t\\0-17.3=-9.18\,t\\t=\frac{17.3}{9.18} \,s\\t\approx 1.88\,s[/tex]

Now, the space covered during these 1.88 s when the vehicle reaches full stop while it decelerates is calculated via:

[tex]x_f-x_i=v_i\,t + \frac{1}{2} a\,t^2\\x_f-x_i=17.3\,t-\frac{9.18}{2} \,t^2\\x_f-x_i=17.3\,(1.88)-4.59\,(1.88)^2\\x_f-x_i=16.3\,ft[/tex]

So, the maximum amount of time the ranger has to react and press the break while driving at 17.3 ft/s is the time to cover 242 ft minus 16.3 ft = 225.7 ft

During 225.7 ft the ranger could be driving in uniform motion (with speed 17.3 ft per second), we find the time to cover such:

[tex]x_f-x_i=v_i\,t\\225.7 = 17.3\,t\\t= \frac{225.7}{17.3} \\t\approx 13\, seconds[/tex]