Explanation:
Given that,
Initial speed of a ranger, u = 36.2 mi/h
Distance dove by the ranger, d = 205 ft
Due to the application of brakes, the acceleration reached is 8.83 ft/s².
We need to find the maximum reaction time allowed if she is to avoid hitting the deer.
We know that,
1 mph = 1.46667 ft/s
36.2 mi/h = 53.09 ft/s
Let t is time.
Using second equation of kinematics to find it as follows :
[tex]d=ut+\dfrac{1}{2}at^2\\\\ 205=53.09t-\dfrac{1}{2}\times 8.83t^2\\\\4.415t^{2}-53.09t+205=0[/tex]
The above is a quadratic equation. We need to solve it for t as follows :
[tex]t=x=\dfrac{-\left(-53.09\right)-4\left(4.415\right)\left(205\right)}{2\cdot4.415},\dfrac{-\left(-53.09\right)+4\left(4.415\right)\left(205\right)}{2\cdot4.415}\\\\t=-403.98\ s,416.01\ s[/tex]
Hence, 416.01 seconds is the maximum reaction time allowed if she is to avoid hitting the deer.
