A ranger in a national park is driving at 34.0 mi/h when a deer jumps into the road 215 ft ahead of the vehicl?

e After a reaction time t, the ranger applies the brakes to produce an acceleration of a = -9.00 ft/s2. What is the maximum reaction time allowed if she is to avoid hitting the deer?A ranger in a national park is driving at 34.0 mi/h when a deer jumps into the road 215 ft ahead of the vehicl?
Below is an answer I already answered for someone else. Just plug your numbers in and you should get your answer.











a= acceleration (-7 m/s^)(-22.9658 fps^)


g= 32.2 fps^


d= distance (50 m)(164.0419 feet)


Vi= initial velocity (15m/s)(49.2125 fps)


s= speed after converting velocity





Here we go.





You need to determine the coefficient (f) of the road. That is done by the following:





f=a/g





Then you figure how far (d) it would take to stop at max braking:





d=s^/30(f)





Then subtract the skid distance from the initial sight distance. Divide the dividend by the initial velocity and you how much time elapses between the initial sighting of the deer and the maximum time the driver could wait to hit the brakes.





I came up with 2.2582 seconds