Question

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.

Answer 1

Answer:

The maximum reaction time is approximately 13 seconds

Explanation:

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:

$11.8\, \frac{mi}{h} =11.8 \,\frac{5280\,ft}{3600\,s} \approx 17.3\,\frac{ft}{s}$

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):

$v_f-v_i=-9.18 \,t\\0-17.3=-9.18\,t\\t=\frac{17.3}{9.18} \,s\\t\approx 1.88\,s$

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

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

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:

$x_f-x_i=v_i\,t\\225.7 = 17.3\,t\\t= \frac{225.7}{17.3} \\t\approx 13\, seconds$