Question

A ranger in a national park is driving at 28.0 mi/h when a deer jumps into the road 155 ft ahead of the vehicle. 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?

Answer 1

1. A ranger in a national park is driving at $28.0 \, \mathrm{mi/h}$ when a deer jumps into the road 155 ft ahead of the vehicle. After a reaction time $t$, the ranger applies the brakes to produce an acceleration of $a = -9.00 \, \mathrm{ft/s^2}$. What is the maximum reaction time allowed if she is to avoid hitting the deer?

**Solution.**

The maximum reaction time is for the case when the ranger stopped directly opposite the deer, so he covers the distance $d$. The movement of the ranger's car is with the constant velocity during the time $t$, and then with the constant acceleration until stopping the car:

$$d = v \cdot t + \frac{v^2}{2a}.$$

We can express the time from the last equation:

$$t = \frac{d}{v} - \frac{v}{2a}.$$

Let check the dimension.

$$[t] = \begin{array}{l} \frac{ft}{ft} - \frac{ft}{s} \\ \frac{ft}{s} - \frac{ft}{s^2} = s. \end{array}$$

Let evaluate the quantity.

$$t = \frac{155}{41.1} - \frac{41.1}{2 \cdot 9} = 1.49(s).$$

**Answer:** $1.49s$.