Question

A ranger in a national park is driving at 52 km/h when a deer jumps onto the road 87 m ahead of the vehicle. After a reaction time of t s, the ranger applies the brakes to produce a deceleration of 4.0 m/s2 . What is the maximum reaction time allowed if the ranger is to avoid hitting the deer? Answer in units of s.

Answer 1

Answer:

Time, t = 0.23 seconds

Explanation:

It is given that,

Initial speed of the ranger, u = 52 km/h = 14.44 m/s

Final speed of the ranger, v = 0 (as brakes are applied)

Acceleration of the ranger, $a=-4\ m/s^2$

Distance between deer and the vehicle, d = 87 m

Let d' is the distance covered by the deer so that it comes top rest. So,

$d'=\dfrac{v^2-u^2}{2a}$

$d'=\dfrac{-(14.44)^2}{2\times -4}$

d' = 26.06 m

Distance between the point where the deer stops and the vehicle is :

D=d-d'

D=87 - 26.06 = 60.94 m

Let t is the maximum reaction time allowed if the ranger is to avoid hitting the deer. It can be calculated as :

$t=\dfrac{v}{D}$

$t=\dfrac{14.44}{60.94}$

t = 0.23 seconds

Hence, this is the required solution.