# A motorist traveling at 17 m/s encounters a deer in the road 39 m ahead. If the maximum acceleration the vehicle’s brakes are capable of is

A motorist traveling at 17 m/s encounters a deer in the road 39 m ahead. If the maximum acceleration the vehicle’s brakes are capable of is −7 m/s 2 , what is the maximum reaction time of the motorist that will allow her or him to avoid hitting the deer? Answer in units of s. 019 (part 2 of 2) 10.0 points If his or her reaction time is 1.21983 s, how fast will (s) he be traveling when (s)he reaches the deer? Answer in units of m/s.

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1. thongdat2

1.     t_reaction = 1.08 s

2.     v₀₁ = 16.365 m/s

Explanation:

1. This is a kinematics exercise, let’s analyze the situation a bit, we can calculate the braking distance and the rest of the distance we can use to calculate the reaction time.

Braking distance

v² = v₀² + 2 a x

when he finishes braking the speed is v = 0

0 = v₀² + 2 a x

x = -v₀² / 2a

x = – 17²/2 (-7)

x = 20.64 m

the distance for the reaction is

d = x_reaction + x

x_reaction = d – x

x_reaction = 39 – 20.64

x_reaction = 18.36 m

as long as it has not reacted the vehicle speed is constant

v = x_reaction / t_reaction

t_reaction = x_reaction / v

t_reaction = 18.36 / 17

t_reaction = 1.08 s

2. Let’s find the distance traveled in the reaction time of t1 = 1.21983 s

as the speed is constant

v = x / t

x₁ = v t₁

the distance traveled during braking is

v² = v₀² + 2a x₂

0 = v₀² + 2 a x₂

x₂ = -v₀² / 2a

v = v₀

the total distance is

x_total = x₁ + x₂

x_total = v₀ t₁ + v₀² / 2a

39 = v₀ 1.21983 + v₀²/14

v₀² + 17.08 vo – 546 =0

we solve the second degree equation

v₀ = [ -17.08 ±√(17.08²  + 4  546) ]/2

v₀ = [-17.08 ± 49.81 ]/2

v₀₁ = 16.365 m/s

v₀₂ = – 33.445 m/s

as the acceleration is negative the correct result is v₀₁ = 16.365 m/s