Question

A ranger in a national park is driving at 36.2mi / h when a deer jumps into the road 205 ft ahead of the vehicle. After a reac tion time of t the ranger applies the brakes to produce and acceleration of - 8.83ft / (s ^ 2) What is the maximum reaction time al lowed if she is to avoid hitting the deer ? Answer in units of s. Need help ASAP please

Answer 1

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 :

$d=ut+\dfrac{1}{2}at^2\\\\ 205=53.09t-\dfrac{1}{2}\times 8.83t^2\\\\4.415t^{2}-53.09t+205=0$

The above is a quadratic equation. We need to solve it for t as follows :

$t=x=\dfrac{-\left(-53.09\right)-4\left(4.415\right)\left(205\right)}{2\cdot4.415},\dfrac{-\left(-53.09\right)+4\left(4.415\right)\left(205\right)}{2\cdot4.415}\\\\t=-403.98\ s,416.01\ s$

Hence, 416.01 seconds is the maximum reaction time allowed if she is to avoid hitting the deer.