Question

A ranger in a national park is driving at 56 km/h when a deer jumps onto the road 65 m ahead of the vehicle. After a reaction time of t s, the ranger applies the brakes to produce an acceleration of â3.0 m/s². What is the maximum reaction time allowed if the ranger is to avoid hitting the deer?

Answer 1

Answer:

 t = 1.58 s

Explanation:

given,

Speed of ranger, v = 56 km/h

                            v = 56 x 0.278 = 15.57 m/s

distance, d = 65 m

deceleration,a = 3 m/s²

reaction time = ?

using stopping distance formula

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

$t = \dfrac{d}{v} -\dfrac{v}{2a}$

t is the reaction time

$t = \dfrac{65}{15.57} -\dfrac{15.57}{2\times 3}$

 t = 1.58 s

hence, the reaction time of the ranger is equal to 1.58 s.